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Find the general solution of the d.e $left(4x^2-xright)y''+2left(2x-1right)y'-4y=12x^2-6x$ given $y_1=frac1x$
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Verify that $y_1=frac1x$ is a solution to d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$
Find the general solution of the d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=frac1x$ however confused on second part.
differential-equations
edited Jul 31 at 16:20
Batominovski
22.8k22776
asked Jul 31 at 15:13
Liz
446
add a comment |Â
up vote
3
down vote
favorite
Verify that $y_1=frac1x$ is a solution to d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$
Find the general solution of the d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=frac1x$ however confused on second part.
differential-equations
edited Jul 31 at 16:20
Batominovski
22.8k22776
asked Jul 31 at 15:13
Liz
446
what is $y1$ again ?
– Ahmad Bazzi
Jul 31 at 15:15
1
@Liz: Are you sure you wrote everything correctly?
– Moo
Jul 31 at 15:24
Thanks .. should have checked before posting :)
– Liz
Jul 31 at 15:29
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Verify that $y_1=frac1x$ is a solution to d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$
Find the general solution of the d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=frac1x$ however confused on second part.
differential-equations
edited Jul 31 at 16:20
Batominovski
22.8k22776
asked Jul 31 at 15:13
Liz
446
Verify that $y_1=frac1x$ is a solution to d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$
Find the general solution of the d.e
$left(4x^2-xright)y''+2left(2x-1right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=frac1x$ however confused on second part.
differential-equations
edited Jul 31 at 16:20
Batominovski
22.8k22776
asked Jul 31 at 15:13
Liz
446
edited Jul 31 at 16:20
Batominovski
22.8k22776
edited Jul 31 at 16:20
Batominovski
22.8k22776
edited Jul 31 at 16:20
Batominovski
22.8k22776
22.8k22776
asked Jul 31 at 15:13
Liz
446
asked Jul 31 at 15:13
Liz
446
asked Jul 31 at 15:13
Liz
446
446
what is $y1$ again ?
– Ahmad Bazzi
Jul 31 at 15:15
1
@Liz: Are you sure you wrote everything correctly?
– Moo
Jul 31 at 15:24
Thanks .. should have checked before posting :)
– Liz
Jul 31 at 15:29
add a comment |Â
what is $y1$ again ?
– Ahmad Bazzi
Jul 31 at 15:15
1
@Liz: Are you sure you wrote everything correctly?
– Moo
Jul 31 at 15:24
Thanks .. should have checked before posting :)
– Liz
Jul 31 at 15:29
what is $y1$ again ?
– Ahmad Bazzi
Jul 31 at 15:15
what is $y1$ again ?
– Ahmad Bazzi
Jul 31 at 15:15
1
1
@Liz: Are you sure you wrote everything correctly?
– Moo
Jul 31 at 15:24
@Liz: Are you sure you wrote everything correctly?
– Moo
Jul 31 at 15:24
Thanks .. should have checked before posting :)
– Liz
Jul 31 at 15:29
Thanks .. should have checked before posting :)
– Liz
Jul 31 at 15:29
add a comment |Â
6 Answers
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up vote
1
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Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + frac4x-2x(4x-1)y' - frac4x(4x-1)y = frac12x-64x-1 $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= frac1xu(x) $$
Plugging this in, we obtain
$$ u'' + left(-frac2x + frac4x-2x(4x-1) right)u' = frac12x^2-6x4x-1 $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working
beginalign u'' - frac44x-1u' &= frac12x^2-6x4x-1 \ fracu''4x-1 - frac4u'(4x-1)^2 &= frac12x^2-6x(4x-1)^2 \ left(fracu'4x-1right)' &= frac34 - frac3/4(4x-1)^2 \ fracu'4x-1 &= frac3x4 + frac3/164x-1 + C \ u' &= 3x^2 + left(C-frac316right)(4x-1) \ u(x) &= x^3 + c_1(2x^2-x) + c_2 \ y(x) &= x^2 + c_1(2x-1) + fracc_2x endalign
answered Jul 31 at 18:38
Dylan
11.4k31026
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
add a comment |Â
up vote
1
down vote
Remark. I need to write a separate answer for the other part of the question because MathJax is disturbingly slow. However, I shall use notations from my first post. Hence, please read that answer first.
You can use the Wronskian technique as mrtaurho suggests. A Wronskian $W(x)$ of the differential equation is
$$W(x)=detleft(beginbmatrix upsilon_1(x)&upsilon_2(x)\ upsilon_1'(x)&upsilon_2'(x)endbmatrixright)=-frac8(4x-1)x^2,,$$
where $upsilon_1(x):=dfrac(4x-1)^2x$ and $upsilon_2(x):=dfrac1x$ are linearly independent solutions I found in my first answer. Then, a particular solution $y:=y_p$ to
$$y''(x)-u(x),y'(x)+v(x),y(x)=frac6(2x-1)4x-1=:t(x)tag$Box$$$
is given by
$$y_p(x)=-upsilon_1(x),int,fracupsilon_2(x),t(x)W(x),textdx+upsilon_2(x),int,fracupsilon_1(x),t(x)W(x),textdx,.$$
That is, a solution is $$beginaligny_p(x) &=frac3(4x-1)^24x,int,fracx(2x-1)(4x-1)^2,textdx-frac34x,int,x(2x-1),textdx \ &=frac3(4x-1)^24x,left(frac8x^2-4x+116(4x-1)+frac116right)-frac34x,left(frac2x^33-fracx^22right) \ &=frac3x(4x-1)8-fracx(4x-3)8=x^2 ,.endalign$$ Thus, all solutions to ($Box$) take the form $$beginalign y(x)&=y_p(x)+A,upsilon_1(x)+B,upsilon_2(x) \ &=x^2+A,left(frac(4x-1)^2xright)+B,left(frac1xright)=x^2+A',(2x-1)+fracB'x,,endalign$$ for some constants $A,A',B,B'$.
Alternatively, you can solve for a particular solution $y:=y_p$ to
$$big(D-f(X)big),big(D-g(X)big),y=t,.$$
By setting $z_p:=big(D-g(X)big),y_p$, we get a solution
$$z_p(x)=frac1nu(x),int,nu(x),t(x),textdx,,text with nu(x):=expleft(-int,f(x),textdxright)=x(4x-1),.$$
By picking an appropriate integral constant, we have
$$z_p(x)=frac1x(4x-1),int,6x(2x-1),textdx=fracx^2(4x-3)x(4x-1)=fracx(4x-3)4x-1,.$$
Then, we solve for a solution $y_p$ to $big(D-g(X)big),y_p=z_p$. We get that $$y_p(x)=frac1mu(x),int,mu(x),z(x),textdx=frac(4x-1)^2x,int,fracx(4x-3)(4x-1)^3,textdx,,$$ where $mu(x)=dfracx(4x-1)^2$ is as before. Ergo, picking one integral constant, we obtain a particular solution $y_p$ to the nonhomogeneous differential equation ($Box$): $$y_p(x)=frac(4x-1)^2x,left(fracx^3(4x-1)^2right)=x^2,.$$
answered Jul 31 at 19:10
Batominovski
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If you have a solution of the homogeneous equation associated to the linear equation, you can make a reduction of order on the equation. It is particularly useful here, because we know how to solve a first order linear equation.
If we write the equation in the form
$$ y''+p(x)y+q(x)y = g(x), $$
since we know that a solution of the associated homogeneous linear equation $ y''+p(x)y+q(x)y = 0 $ is $ y_h = 1/x$, we can look for a solution to the non-homogeneous equation of the form $y_s=y_hu$, where $u$ is a function we must determine. Therefore, we have
$$ y_s ' = y_h 'u + y_h,u' , y_s '' = y_h '' u +2y_h ' u' + y_h u'' $$
putting it in the equation leads to
$$ y_h '' u + 2 y_h ' u + y_h u'' + p(y_h ' u + y_h u') + q (y_h u) = g $$
$$ implies u(y_h '' + p y_h ' + qy_h) + 2y_h ' u' + y_h u'' + py_h u' =g $$
$$ implies 2y_h ' u' + y_h u'' + py_h u' = g $$
So we just have to solve this first order linear equation for $u$ to get a solution of the equation $y''+p(x)y+q(x)y = g(x)$. The same can be done to find another linear independent solution to the homogeneous equation.
answered Jul 31 at 18:22
zaphodxvii
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HINT
First of all bring the equation in the form
$$y''+p(x)y'+q(x)y~=~g(x)$$
Use the fact that there are two different ways to compute the Wronskian of a fundamental set of solutions:
$$beginalign
&(1)W(y_1,y_2)~=~beginvmatrixy_1&y_2\y_1'&y_2'endvmatrix~=~y_1y_2'-y_2y_1'\
&(2)W(y_1,y_2)~=~e^-int p(x)dx
endalign$$
From thereone you can compute the general solution to the homogenous equation by solving the ODE for $y_2$. After this you can do variation of constants to determine the solution to your given equation.
answered Jul 31 at 15:41
mrtaurho
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Not an answer just a hint
$$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$$
Assume that $y_2=P(x)$ is a solution. Where
$$P(x)=sum_n=0^ma_nx^n$$
Whats the degree of $P(x)$ ? plug the solution you get that
$$a_nn(n-1)x^n=0 ,, forall x implies n=0, n=1 $$
For $n=0$ you get the trivial solution $y=0$
Try $n=1$
$$ y_2=ax+b$$
$$2left(2x-1right)a-4(ax+b)=0$$
$$implies a=-2b$$
Therefore
$$y_2=-2bx+b=b(1-2x)$$
Therefore the solution for the homogeneous equation is
$$y_h=frac c_1x+c_2(1-2x)$$
Use variation of constant method to find $y_p$
answered Jul 31 at 17:12
Isham
10.5k3829
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Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
1
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
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Warning. This is a very wild solution for the first part of the problem (i.e., solving the homogeneous differential equation), intended only to illustrate its beauty (if you would agree), and not intended to be practical. This solution hinges on some heavy guess work, but it does not assume to know any solution to the required differential equaiton. That is, I was looking for a solution $g$ to the differential equation
$$g'(x)+big(g(x)big)^2-u(x),g(x)+v(x)=0,,tag$star$$$
where $u(x):=-dfrac2(2x-1)x(4x-1)$ and $v(x):=-dfrac4x(4x-1)$. I guessed that $g(x)=dfracax+bx(4x-1)$ for some $a$ and $b$, and then I solved for $a$ and $b$. I shall omit the gruesome detail of how I obtained $a$ and $b$, but it is really nothing but expanding terms in ($star$).
Let
$$f(x):=-frac8x-1x(4x-1)text and g(x):=frac4x+1x(4x-1),.$$ Observe that
$$f(x)+g(x)=u(x)text and f(x),g(x)-g'(x)=v(x),.$$
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2-u(X),D+v(X)=big(D-f(X)big),big(D-g(X)big),.tag*$$
Suppose that a function $y$ satisfies $$big(D^2-u(X),D+v(X)big),y=0,.tag#$$ Let $z:=big(D-g(X)big),y$. By (*) and (#),
$$big(D-f(X)big),z=0,.$$
Consequently,
$$z'(x)-f(x),z(x)=0text or z(x)=expleft(int,f(x),textdxright),.$$
This is a not-so-difficult work, and the result is
$$z(x)=frackx(4x-1)text for some constant k,.$$
Now, we want to solve $big(D-g(X)big),y=z$, or $$y'(x)-g(x),y(x)=z(x),.$$ That is, $$y(x)=frac1mu(x),int,mu(x),z(x),textdx,,text with mu(x):=expleft(-int,g(x),textdxright),.$$ We see that $mu(x)=dfracx(4x-1)^2$. Therefore, $$y(x)=frac(4x-1)^2x,int,fracx(4x-1)^2,left(frackx(4x-1)right),textdx,.$$ Ergo, $$y(x)=frac(4x-1)^2x,Biggl(A+B,left(frac1(4x-1)^2right)Biggr),,$$ where $A$ is a constant and $B:=-dfrack8$. This means $$y(x)=A,left(frac(4x-1)^2xright)+Bleft(frac1xright),.$$
Thus, the given solution $y_1(x)=dfrac1x$ corresponds to $(A,B):=(0,1)$. The linear solution $y_2(x)=2x-1$ found by Isham comes from $(A,B):=left(dfrac18,-dfrac18right)$.
P.S. This solution is inspired by a method of solving the quantum harmonic oscillator problem. One such method writes $$D^2-X^2+I=(D-X),(D+X),,$$
where $I$ is the identity operator.
answered Jul 31 at 18:32
Batominovski
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6 Answers
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6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + frac4x-2x(4x-1)y' - frac4x(4x-1)y = frac12x-64x-1 $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= frac1xu(x) $$
Plugging this in, we obtain
$$ u'' + left(-frac2x + frac4x-2x(4x-1) right)u' = frac12x^2-6x4x-1 $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working
beginalign u'' - frac44x-1u' &= frac12x^2-6x4x-1 \ fracu''4x-1 - frac4u'(4x-1)^2 &= frac12x^2-6x(4x-1)^2 \ left(fracu'4x-1right)' &= frac34 - frac3/4(4x-1)^2 \ fracu'4x-1 &= frac3x4 + frac3/164x-1 + C \ u' &= 3x^2 + left(C-frac316right)(4x-1) \ u(x) &= x^3 + c_1(2x^2-x) + c_2 \ y(x) &= x^2 + c_1(2x-1) + fracc_2x endalign
answered Jul 31 at 18:38
Dylan
11.4k31026
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
add a comment |Â
up vote
1
down vote
Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + frac4x-2x(4x-1)y' - frac4x(4x-1)y = frac12x-64x-1 $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= frac1xu(x) $$
Plugging this in, we obtain
$$ u'' + left(-frac2x + frac4x-2x(4x-1) right)u' = frac12x^2-6x4x-1 $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working
beginalign u'' - frac44x-1u' &= frac12x^2-6x4x-1 \ fracu''4x-1 - frac4u'(4x-1)^2 &= frac12x^2-6x(4x-1)^2 \ left(fracu'4x-1right)' &= frac34 - frac3/4(4x-1)^2 \ fracu'4x-1 &= frac3x4 + frac3/164x-1 + C \ u' &= 3x^2 + left(C-frac316right)(4x-1) \ u(x) &= x^3 + c_1(2x^2-x) + c_2 \ y(x) &= x^2 + c_1(2x-1) + fracc_2x endalign
answered Jul 31 at 18:38
Dylan
11.4k31026
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + frac4x-2x(4x-1)y' - frac4x(4x-1)y = frac12x-64x-1 $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= frac1xu(x) $$
Plugging this in, we obtain
$$ u'' + left(-frac2x + frac4x-2x(4x-1) right)u' = frac12x^2-6x4x-1 $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working
beginalign u'' - frac44x-1u' &= frac12x^2-6x4x-1 \ fracu''4x-1 - frac4u'(4x-1)^2 &= frac12x^2-6x(4x-1)^2 \ left(fracu'4x-1right)' &= frac34 - frac3/4(4x-1)^2 \ fracu'4x-1 &= frac3x4 + frac3/164x-1 + C \ u' &= 3x^2 + left(C-frac316right)(4x-1) \ u(x) &= x^3 + c_1(2x^2-x) + c_2 \ y(x) &= x^2 + c_1(2x-1) + fracc_2x endalign
answered Jul 31 at 18:38
Dylan
11.4k31026
Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + frac4x-2x(4x-1)y' - frac4x(4x-1)y = frac12x-64x-1 $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= frac1xu(x) $$
Plugging this in, we obtain
$$ u'' + left(-frac2x + frac4x-2x(4x-1) right)u' = frac12x^2-6x4x-1 $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working
beginalign u'' - frac44x-1u' &= frac12x^2-6x4x-1 \ fracu''4x-1 - frac4u'(4x-1)^2 &= frac12x^2-6x(4x-1)^2 \ left(fracu'4x-1right)' &= frac34 - frac3/4(4x-1)^2 \ fracu'4x-1 &= frac3x4 + frac3/164x-1 + C \ u' &= 3x^2 + left(C-frac316right)(4x-1) \ u(x) &= x^3 + c_1(2x^2-x) + c_2 \ y(x) &= x^2 + c_1(2x-1) + fracc_2x endalign
answered Jul 31 at 18:38
Dylan
11.4k31026
edited Jul 31 at 19:37
answered Jul 31 at 18:38
Dylan
11.4k31026
answered Jul 31 at 18:38
Dylan
11.4k31026
answered Jul 31 at 18:38
Dylan
11.4k31026
11.4k31026
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
add a comment |Â
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
HI Dylan, is there a name for this particular form of general solution that you have stated. Just seeing if I can read up a little more on it. Thanks again soln was easy to follow :)
– Liz
Jul 31 at 21:25
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
It's just the form of the solution used in the method of reduction of order. You can read up on that method.
– Dylan
Jul 31 at 21:49
add a comment |Â
up vote
1
down vote
Remark. I need to write a separate answer for the other part of the question because MathJax is disturbingly slow. However, I shall use notations from my first post. Hence, please read that answer first.
You can use the Wronskian technique as mrtaurho suggests. A Wronskian $W(x)$ of the differential equation is
$$W(x)=detleft(beginbmatrix upsilon_1(x)&upsilon_2(x)\ upsilon_1'(x)&upsilon_2'(x)endbmatrixright)=-frac8(4x-1)x^2,,$$
where $upsilon_1(x):=dfrac(4x-1)^2x$ and $upsilon_2(x):=dfrac1x$ are linearly independent solutions I found in my first answer. Then, a particular solution $y:=y_p$ to
$$y''(x)-u(x),y'(x)+v(x),y(x)=frac6(2x-1)4x-1=:t(x)tag$Box$$$
is given by
$$y_p(x)=-upsilon_1(x),int,fracupsilon_2(x),t(x)W(x),textdx+upsilon_2(x),int,fracupsilon_1(x),t(x)W(x),textdx,.$$
That is, a solution is $$beginaligny_p(x) &=frac3(4x-1)^24x,int,fracx(2x-1)(4x-1)^2,textdx-frac34x,int,x(2x-1),textdx \ &=frac3(4x-1)^24x,left(frac8x^2-4x+116(4x-1)+frac116right)-frac34x,left(frac2x^33-fracx^22right) \ &=frac3x(4x-1)8-fracx(4x-3)8=x^2 ,.endalign$$ Thus, all solutions to ($Box$) take the form $$beginalign y(x)&=y_p(x)+A,upsilon_1(x)+B,upsilon_2(x) \ &=x^2+A,left(frac(4x-1)^2xright)+B,left(frac1xright)=x^2+A',(2x-1)+fracB'x,,endalign$$ for some constants $A,A',B,B'$.
Alternatively, you can solve for a particular solution $y:=y_p$ to
$$big(D-f(X)big),big(D-g(X)big),y=t,.$$
By setting $z_p:=big(D-g(X)big),y_p$, we get a solution
$$z_p(x)=frac1nu(x),int,nu(x),t(x),textdx,,text with nu(x):=expleft(-int,f(x),textdxright)=x(4x-1),.$$
By picking an appropriate integral constant, we have
$$z_p(x)=frac1x(4x-1),int,6x(2x-1),textdx=fracx^2(4x-3)x(4x-1)=fracx(4x-3)4x-1,.$$
Then, we solve for a solution $y_p$ to $big(D-g(X)big),y_p=z_p$. We get that $$y_p(x)=frac1mu(x),int,mu(x),z(x),textdx=frac(4x-1)^2x,int,fracx(4x-3)(4x-1)^3,textdx,,$$ where $mu(x)=dfracx(4x-1)^2$ is as before. Ergo, picking one integral constant, we obtain a particular solution $y_p$ to the nonhomogeneous differential equation ($Box$): $$y_p(x)=frac(4x-1)^2x,left(fracx^3(4x-1)^2right)=x^2,.$$
answered Jul 31 at 19:10
Batominovski
22.8k22776
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up vote
1
down vote
Remark. I need to write a separate answer for the other part of the question because MathJax is disturbingly slow. However, I shall use notations from my first post. Hence, please read that answer first.
You can use the Wronskian technique as mrtaurho suggests. A Wronskian $W(x)$ of the differential equation is
$$W(x)=detleft(beginbmatrix upsilon_1(x)&upsilon_2(x)\ upsilon_1'(x)&upsilon_2'(x)endbmatrixright)=-frac8(4x-1)x^2,,$$
where $upsilon_1(x):=dfrac(4x-1)^2x$ and $upsilon_2(x):=dfrac1x$ are linearly independent solutions I found in my first answer. Then, a particular solution $y:=y_p$ to
$$y''(x)-u(x),y'(x)+v(x),y(x)=frac6(2x-1)4x-1=:t(x)tag$Box$$$
is given by
$$y_p(x)=-upsilon_1(x),int,fracupsilon_2(x),t(x)W(x),textdx+upsilon_2(x),int,fracupsilon_1(x),t(x)W(x),textdx,.$$
That is, a solution is $$beginaligny_p(x) &=frac3(4x-1)^24x,int,fracx(2x-1)(4x-1)^2,textdx-frac34x,int,x(2x-1),textdx \ &=frac3(4x-1)^24x,left(frac8x^2-4x+116(4x-1)+frac116right)-frac34x,left(frac2x^33-fracx^22right) \ &=frac3x(4x-1)8-fracx(4x-3)8=x^2 ,.endalign$$ Thus, all solutions to ($Box$) take the form $$beginalign y(x)&=y_p(x)+A,upsilon_1(x)+B,upsilon_2(x) \ &=x^2+A,left(frac(4x-1)^2xright)+B,left(frac1xright)=x^2+A',(2x-1)+fracB'x,,endalign$$ for some constants $A,A',B,B'$.
Alternatively, you can solve for a particular solution $y:=y_p$ to
$$big(D-f(X)big),big(D-g(X)big),y=t,.$$
By setting $z_p:=big(D-g(X)big),y_p$, we get a solution
$$z_p(x)=frac1nu(x),int,nu(x),t(x),textdx,,text with nu(x):=expleft(-int,f(x),textdxright)=x(4x-1),.$$
By picking an appropriate integral constant, we have
$$z_p(x)=frac1x(4x-1),int,6x(2x-1),textdx=fracx^2(4x-3)x(4x-1)=fracx(4x-3)4x-1,.$$
Then, we solve for a solution $y_p$ to $big(D-g(X)big),y_p=z_p$. We get that $$y_p(x)=frac1mu(x),int,mu(x),z(x),textdx=frac(4x-1)^2x,int,fracx(4x-3)(4x-1)^3,textdx,,$$ where $mu(x)=dfracx(4x-1)^2$ is as before. Ergo, picking one integral constant, we obtain a particular solution $y_p$ to the nonhomogeneous differential equation ($Box$): $$y_p(x)=frac(4x-1)^2x,left(fracx^3(4x-1)^2right)=x^2,.$$
answered Jul 31 at 19:10
Batominovski
22.8k22776
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up vote
1
down vote
up vote
1
down vote
Remark. I need to write a separate answer for the other part of the question because MathJax is disturbingly slow. However, I shall use notations from my first post. Hence, please read that answer first.
You can use the Wronskian technique as mrtaurho suggests. A Wronskian $W(x)$ of the differential equation is
$$W(x)=detleft(beginbmatrix upsilon_1(x)&upsilon_2(x)\ upsilon_1'(x)&upsilon_2'(x)endbmatrixright)=-frac8(4x-1)x^2,,$$
where $upsilon_1(x):=dfrac(4x-1)^2x$ and $upsilon_2(x):=dfrac1x$ are linearly independent solutions I found in my first answer. Then, a particular solution $y:=y_p$ to
$$y''(x)-u(x),y'(x)+v(x),y(x)=frac6(2x-1)4x-1=:t(x)tag$Box$$$
is given by
$$y_p(x)=-upsilon_1(x),int,fracupsilon_2(x),t(x)W(x),textdx+upsilon_2(x),int,fracupsilon_1(x),t(x)W(x),textdx,.$$
That is, a solution is $$beginaligny_p(x) &=frac3(4x-1)^24x,int,fracx(2x-1)(4x-1)^2,textdx-frac34x,int,x(2x-1),textdx \ &=frac3(4x-1)^24x,left(frac8x^2-4x+116(4x-1)+frac116right)-frac34x,left(frac2x^33-fracx^22right) \ &=frac3x(4x-1)8-fracx(4x-3)8=x^2 ,.endalign$$ Thus, all solutions to ($Box$) take the form $$beginalign y(x)&=y_p(x)+A,upsilon_1(x)+B,upsilon_2(x) \ &=x^2+A,left(frac(4x-1)^2xright)+B,left(frac1xright)=x^2+A',(2x-1)+fracB'x,,endalign$$ for some constants $A,A',B,B'$.
Alternatively, you can solve for a particular solution $y:=y_p$ to
$$big(D-f(X)big),big(D-g(X)big),y=t,.$$
By setting $z_p:=big(D-g(X)big),y_p$, we get a solution
$$z_p(x)=frac1nu(x),int,nu(x),t(x),textdx,,text with nu(x):=expleft(-int,f(x),textdxright)=x(4x-1),.$$
By picking an appropriate integral constant, we have
$$z_p(x)=frac1x(4x-1),int,6x(2x-1),textdx=fracx^2(4x-3)x(4x-1)=fracx(4x-3)4x-1,.$$
Then, we solve for a solution $y_p$ to $big(D-g(X)big),y_p=z_p$. We get that $$y_p(x)=frac1mu(x),int,mu(x),z(x),textdx=frac(4x-1)^2x,int,fracx(4x-3)(4x-1)^3,textdx,,$$ where $mu(x)=dfracx(4x-1)^2$ is as before. Ergo, picking one integral constant, we obtain a particular solution $y_p$ to the nonhomogeneous differential equation ($Box$): $$y_p(x)=frac(4x-1)^2x,left(fracx^3(4x-1)^2right)=x^2,.$$
answered Jul 31 at 19:10
Batominovski
22.8k22776
Remark. I need to write a separate answer for the other part of the question because MathJax is disturbingly slow. However, I shall use notations from my first post. Hence, please read that answer first.
You can use the Wronskian technique as mrtaurho suggests. A Wronskian $W(x)$ of the differential equation is
$$W(x)=detleft(beginbmatrix upsilon_1(x)&upsilon_2(x)\ upsilon_1'(x)&upsilon_2'(x)endbmatrixright)=-frac8(4x-1)x^2,,$$
where $upsilon_1(x):=dfrac(4x-1)^2x$ and $upsilon_2(x):=dfrac1x$ are linearly independent solutions I found in my first answer. Then, a particular solution $y:=y_p$ to
$$y''(x)-u(x),y'(x)+v(x),y(x)=frac6(2x-1)4x-1=:t(x)tag$Box$$$
is given by
$$y_p(x)=-upsilon_1(x),int,fracupsilon_2(x),t(x)W(x),textdx+upsilon_2(x),int,fracupsilon_1(x),t(x)W(x),textdx,.$$
That is, a solution is $$beginaligny_p(x) &=frac3(4x-1)^24x,int,fracx(2x-1)(4x-1)^2,textdx-frac34x,int,x(2x-1),textdx \ &=frac3(4x-1)^24x,left(frac8x^2-4x+116(4x-1)+frac116right)-frac34x,left(frac2x^33-fracx^22right) \ &=frac3x(4x-1)8-fracx(4x-3)8=x^2 ,.endalign$$ Thus, all solutions to ($Box$) take the form $$beginalign y(x)&=y_p(x)+A,upsilon_1(x)+B,upsilon_2(x) \ &=x^2+A,left(frac(4x-1)^2xright)+B,left(frac1xright)=x^2+A',(2x-1)+fracB'x,,endalign$$ for some constants $A,A',B,B'$.
Alternatively, you can solve for a particular solution $y:=y_p$ to
$$big(D-f(X)big),big(D-g(X)big),y=t,.$$
By setting $z_p:=big(D-g(X)big),y_p$, we get a solution
$$z_p(x)=frac1nu(x),int,nu(x),t(x),textdx,,text with nu(x):=expleft(-int,f(x),textdxright)=x(4x-1),.$$
By picking an appropriate integral constant, we have
$$z_p(x)=frac1x(4x-1),int,6x(2x-1),textdx=fracx^2(4x-3)x(4x-1)=fracx(4x-3)4x-1,.$$
Then, we solve for a solution $y_p$ to $big(D-g(X)big),y_p=z_p$. We get that $$y_p(x)=frac1mu(x),int,mu(x),z(x),textdx=frac(4x-1)^2x,int,fracx(4x-3)(4x-1)^3,textdx,,$$ where $mu(x)=dfracx(4x-1)^2$ is as before. Ergo, picking one integral constant, we obtain a particular solution $y_p$ to the nonhomogeneous differential equation ($Box$): $$y_p(x)=frac(4x-1)^2x,left(fracx^3(4x-1)^2right)=x^2,.$$
answered Jul 31 at 19:10
Batominovski
22.8k22776
edited Jul 31 at 20:21
answered Jul 31 at 19:10
Batominovski
22.8k22776
answered Jul 31 at 19:10
Batominovski
22.8k22776
answered Jul 31 at 19:10
Batominovski
22.8k22776
22.8k22776
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1
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If you have a solution of the homogeneous equation associated to the linear equation, you can make a reduction of order on the equation. It is particularly useful here, because we know how to solve a first order linear equation.
If we write the equation in the form
$$ y''+p(x)y+q(x)y = g(x), $$
since we know that a solution of the associated homogeneous linear equation $ y''+p(x)y+q(x)y = 0 $ is $ y_h = 1/x$, we can look for a solution to the non-homogeneous equation of the form $y_s=y_hu$, where $u$ is a function we must determine. Therefore, we have
$$ y_s ' = y_h 'u + y_h,u' , y_s '' = y_h '' u +2y_h ' u' + y_h u'' $$
putting it in the equation leads to
$$ y_h '' u + 2 y_h ' u + y_h u'' + p(y_h ' u + y_h u') + q (y_h u) = g $$
$$ implies u(y_h '' + p y_h ' + qy_h) + 2y_h ' u' + y_h u'' + py_h u' =g $$
$$ implies 2y_h ' u' + y_h u'' + py_h u' = g $$
So we just have to solve this first order linear equation for $u$ to get a solution of the equation $y''+p(x)y+q(x)y = g(x)$. The same can be done to find another linear independent solution to the homogeneous equation.
answered Jul 31 at 18:22
zaphodxvii
1068
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up vote
1
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If you have a solution of the homogeneous equation associated to the linear equation, you can make a reduction of order on the equation. It is particularly useful here, because we know how to solve a first order linear equation.
If we write the equation in the form
$$ y''+p(x)y+q(x)y = g(x), $$
since we know that a solution of the associated homogeneous linear equation $ y''+p(x)y+q(x)y = 0 $ is $ y_h = 1/x$, we can look for a solution to the non-homogeneous equation of the form $y_s=y_hu$, where $u$ is a function we must determine. Therefore, we have
$$ y_s ' = y_h 'u + y_h,u' , y_s '' = y_h '' u +2y_h ' u' + y_h u'' $$
putting it in the equation leads to
$$ y_h '' u + 2 y_h ' u + y_h u'' + p(y_h ' u + y_h u') + q (y_h u) = g $$
$$ implies u(y_h '' + p y_h ' + qy_h) + 2y_h ' u' + y_h u'' + py_h u' =g $$
$$ implies 2y_h ' u' + y_h u'' + py_h u' = g $$
So we just have to solve this first order linear equation for $u$ to get a solution of the equation $y''+p(x)y+q(x)y = g(x)$. The same can be done to find another linear independent solution to the homogeneous equation.
answered Jul 31 at 18:22
zaphodxvii
1068
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you have a solution of the homogeneous equation associated to the linear equation, you can make a reduction of order on the equation. It is particularly useful here, because we know how to solve a first order linear equation.
If we write the equation in the form
$$ y''+p(x)y+q(x)y = g(x), $$
since we know that a solution of the associated homogeneous linear equation $ y''+p(x)y+q(x)y = 0 $ is $ y_h = 1/x$, we can look for a solution to the non-homogeneous equation of the form $y_s=y_hu$, where $u$ is a function we must determine. Therefore, we have
$$ y_s ' = y_h 'u + y_h,u' , y_s '' = y_h '' u +2y_h ' u' + y_h u'' $$
putting it in the equation leads to
$$ y_h '' u + 2 y_h ' u + y_h u'' + p(y_h ' u + y_h u') + q (y_h u) = g $$
$$ implies u(y_h '' + p y_h ' + qy_h) + 2y_h ' u' + y_h u'' + py_h u' =g $$
$$ implies 2y_h ' u' + y_h u'' + py_h u' = g $$
So we just have to solve this first order linear equation for $u$ to get a solution of the equation $y''+p(x)y+q(x)y = g(x)$. The same can be done to find another linear independent solution to the homogeneous equation.
answered Jul 31 at 18:22
zaphodxvii
1068
If you have a solution of the homogeneous equation associated to the linear equation, you can make a reduction of order on the equation. It is particularly useful here, because we know how to solve a first order linear equation.
If we write the equation in the form
$$ y''+p(x)y+q(x)y = g(x), $$
since we know that a solution of the associated homogeneous linear equation $ y''+p(x)y+q(x)y = 0 $ is $ y_h = 1/x$, we can look for a solution to the non-homogeneous equation of the form $y_s=y_hu$, where $u$ is a function we must determine. Therefore, we have
$$ y_s ' = y_h 'u + y_h,u' , y_s '' = y_h '' u +2y_h ' u' + y_h u'' $$
putting it in the equation leads to
$$ y_h '' u + 2 y_h ' u + y_h u'' + p(y_h ' u + y_h u') + q (y_h u) = g $$
$$ implies u(y_h '' + p y_h ' + qy_h) + 2y_h ' u' + y_h u'' + py_h u' =g $$
$$ implies 2y_h ' u' + y_h u'' + py_h u' = g $$
So we just have to solve this first order linear equation for $u$ to get a solution of the equation $y''+p(x)y+q(x)y = g(x)$. The same can be done to find another linear independent solution to the homogeneous equation.
answered Jul 31 at 18:22
zaphodxvii
1068
edited Aug 1 at 0:22
answered Jul 31 at 18:22
zaphodxvii
1068
answered Jul 31 at 18:22
zaphodxvii
1068
answered Jul 31 at 18:22
zaphodxvii
1068
1068
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HINT
First of all bring the equation in the form
$$y''+p(x)y'+q(x)y~=~g(x)$$
Use the fact that there are two different ways to compute the Wronskian of a fundamental set of solutions:
$$beginalign
&(1)W(y_1,y_2)~=~beginvmatrixy_1&y_2\y_1'&y_2'endvmatrix~=~y_1y_2'-y_2y_1'\
&(2)W(y_1,y_2)~=~e^-int p(x)dx
endalign$$
From thereone you can compute the general solution to the homogenous equation by solving the ODE for $y_2$. After this you can do variation of constants to determine the solution to your given equation.
answered Jul 31 at 15:41
mrtaurho
660117
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HINT
First of all bring the equation in the form
$$y''+p(x)y'+q(x)y~=~g(x)$$
Use the fact that there are two different ways to compute the Wronskian of a fundamental set of solutions:
$$beginalign
&(1)W(y_1,y_2)~=~beginvmatrixy_1&y_2\y_1'&y_2'endvmatrix~=~y_1y_2'-y_2y_1'\
&(2)W(y_1,y_2)~=~e^-int p(x)dx
endalign$$
From thereone you can compute the general solution to the homogenous equation by solving the ODE for $y_2$. After this you can do variation of constants to determine the solution to your given equation.
answered Jul 31 at 15:41
mrtaurho
660117
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0
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up vote
0
down vote
HINT
First of all bring the equation in the form
$$y''+p(x)y'+q(x)y~=~g(x)$$
Use the fact that there are two different ways to compute the Wronskian of a fundamental set of solutions:
$$beginalign
&(1)W(y_1,y_2)~=~beginvmatrixy_1&y_2\y_1'&y_2'endvmatrix~=~y_1y_2'-y_2y_1'\
&(2)W(y_1,y_2)~=~e^-int p(x)dx
endalign$$
From thereone you can compute the general solution to the homogenous equation by solving the ODE for $y_2$. After this you can do variation of constants to determine the solution to your given equation.
answered Jul 31 at 15:41
mrtaurho
660117
HINT
First of all bring the equation in the form
$$y''+p(x)y'+q(x)y~=~g(x)$$
Use the fact that there are two different ways to compute the Wronskian of a fundamental set of solutions:
$$beginalign
&(1)W(y_1,y_2)~=~beginvmatrixy_1&y_2\y_1'&y_2'endvmatrix~=~y_1y_2'-y_2y_1'\
&(2)W(y_1,y_2)~=~e^-int p(x)dx
endalign$$
From thereone you can compute the general solution to the homogenous equation by solving the ODE for $y_2$. After this you can do variation of constants to determine the solution to your given equation.
answered Jul 31 at 15:41
mrtaurho
660117
answered Jul 31 at 15:41
mrtaurho
660117
answered Jul 31 at 15:41
mrtaurho
660117
answered Jul 31 at 15:41
mrtaurho
660117
660117
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Not an answer just a hint
$$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$$
Assume that $y_2=P(x)$ is a solution. Where
$$P(x)=sum_n=0^ma_nx^n$$
Whats the degree of $P(x)$ ? plug the solution you get that
$$a_nn(n-1)x^n=0 ,, forall x implies n=0, n=1 $$
For $n=0$ you get the trivial solution $y=0$
Try $n=1$
$$ y_2=ax+b$$
$$2left(2x-1right)a-4(ax+b)=0$$
$$implies a=-2b$$
Therefore
$$y_2=-2bx+b=b(1-2x)$$
Therefore the solution for the homogeneous equation is
$$y_h=frac c_1x+c_2(1-2x)$$
Use variation of constant method to find $y_p$
answered Jul 31 at 17:12
Isham
10.5k3829
1
Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
1
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
add a comment |Â
up vote
0
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Not an answer just a hint
$$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$$
Assume that $y_2=P(x)$ is a solution. Where
$$P(x)=sum_n=0^ma_nx^n$$
Whats the degree of $P(x)$ ? plug the solution you get that
$$a_nn(n-1)x^n=0 ,, forall x implies n=0, n=1 $$
For $n=0$ you get the trivial solution $y=0$
Try $n=1$
$$ y_2=ax+b$$
$$2left(2x-1right)a-4(ax+b)=0$$
$$implies a=-2b$$
Therefore
$$y_2=-2bx+b=b(1-2x)$$
Therefore the solution for the homogeneous equation is
$$y_h=frac c_1x+c_2(1-2x)$$
Use variation of constant method to find $y_p$
answered Jul 31 at 17:12
Isham
10.5k3829
1
Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
1
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Not an answer just a hint
$$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$$
Assume that $y_2=P(x)$ is a solution. Where
$$P(x)=sum_n=0^ma_nx^n$$
Whats the degree of $P(x)$ ? plug the solution you get that
$$a_nn(n-1)x^n=0 ,, forall x implies n=0, n=1 $$
For $n=0$ you get the trivial solution $y=0$
Try $n=1$
$$ y_2=ax+b$$
$$2left(2x-1right)a-4(ax+b)=0$$
$$implies a=-2b$$
Therefore
$$y_2=-2bx+b=b(1-2x)$$
Therefore the solution for the homogeneous equation is
$$y_h=frac c_1x+c_2(1-2x)$$
Use variation of constant method to find $y_p$
answered Jul 31 at 17:12
Isham
10.5k3829
Not an answer just a hint
$$left(4x^2-xright)y''+2left(2x-1right)y'-4y=0$$
Assume that $y_2=P(x)$ is a solution. Where
$$P(x)=sum_n=0^ma_nx^n$$
Whats the degree of $P(x)$ ? plug the solution you get that
$$a_nn(n-1)x^n=0 ,, forall x implies n=0, n=1 $$
For $n=0$ you get the trivial solution $y=0$
Try $n=1$
$$ y_2=ax+b$$
$$2left(2x-1right)a-4(ax+b)=0$$
$$implies a=-2b$$
Therefore
$$y_2=-2bx+b=b(1-2x)$$
Therefore the solution for the homogeneous equation is
$$y_h=frac c_1x+c_2(1-2x)$$
Use variation of constant method to find $y_p$
answered Jul 31 at 17:12
Isham
10.5k3829
edited Jul 31 at 17:33
answered Jul 31 at 17:12
Isham
10.5k3829
answered Jul 31 at 17:12
Isham
10.5k3829
answered Jul 31 at 17:12
Isham
10.5k3829
10.5k3829
1
Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
1
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
add a comment |Â
1
Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
1
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
1
1
Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Just an observation: If we don't know anything about the solution, we can use the power series method to obtain the $1-2x$ solution, then apply reduction of order to find the other $1/x$ solution. The given problem makes us work the other way around, i.e. use the $1/x$ solution to find the $1-2x$ solution
– Dylan
Jul 31 at 19:43
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
Yeah @Dylan... I just tried a polynomial solution here rather than using general mehod ( Wronskian...)
– Isham
Jul 31 at 19:50
1
1
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
What I'm saying is, this is a great method if you don't know the solution. The Wronskian method requires that you know one of them.
– Dylan
Jul 31 at 20:54
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
yes I agree @Dylan ...serie solution sometimes works fine
– Isham
Jul 31 at 20:58
add a comment |Â
up vote
0
down vote
Warning. This is a very wild solution for the first part of the problem (i.e., solving the homogeneous differential equation), intended only to illustrate its beauty (if you would agree), and not intended to be practical. This solution hinges on some heavy guess work, but it does not assume to know any solution to the required differential equaiton. That is, I was looking for a solution $g$ to the differential equation
$$g'(x)+big(g(x)big)^2-u(x),g(x)+v(x)=0,,tag$star$$$
where $u(x):=-dfrac2(2x-1)x(4x-1)$ and $v(x):=-dfrac4x(4x-1)$. I guessed that $g(x)=dfracax+bx(4x-1)$ for some $a$ and $b$, and then I solved for $a$ and $b$. I shall omit the gruesome detail of how I obtained $a$ and $b$, but it is really nothing but expanding terms in ($star$).
Let
$$f(x):=-frac8x-1x(4x-1)text and g(x):=frac4x+1x(4x-1),.$$ Observe that
$$f(x)+g(x)=u(x)text and f(x),g(x)-g'(x)=v(x),.$$
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2-u(X),D+v(X)=big(D-f(X)big),big(D-g(X)big),.tag*$$
Suppose that a function $y$ satisfies $$big(D^2-u(X),D+v(X)big),y=0,.tag#$$ Let $z:=big(D-g(X)big),y$. By (*) and (#),
$$big(D-f(X)big),z=0,.$$
Consequently,
$$z'(x)-f(x),z(x)=0text or z(x)=expleft(int,f(x),textdxright),.$$
This is a not-so-difficult work, and the result is
$$z(x)=frackx(4x-1)text for some constant k,.$$
Now, we want to solve $big(D-g(X)big),y=z$, or $$y'(x)-g(x),y(x)=z(x),.$$ That is, $$y(x)=frac1mu(x),int,mu(x),z(x),textdx,,text with mu(x):=expleft(-int,g(x),textdxright),.$$ We see that $mu(x)=dfracx(4x-1)^2$. Therefore, $$y(x)=frac(4x-1)^2x,int,fracx(4x-1)^2,left(frackx(4x-1)right),textdx,.$$ Ergo, $$y(x)=frac(4x-1)^2x,Biggl(A+B,left(frac1(4x-1)^2right)Biggr),,$$ where $A$ is a constant and $B:=-dfrack8$. This means $$y(x)=A,left(frac(4x-1)^2xright)+Bleft(frac1xright),.$$
Thus, the given solution $y_1(x)=dfrac1x$ corresponds to $(A,B):=(0,1)$. The linear solution $y_2(x)=2x-1$ found by Isham comes from $(A,B):=left(dfrac18,-dfrac18right)$.
P.S. This solution is inspired by a method of solving the quantum harmonic oscillator problem. One such method writes $$D^2-X^2+I=(D-X),(D+X),,$$
where $I$ is the identity operator.
answered Jul 31 at 18:32
Batominovski
22.8k22776
add a comment |Â
up vote
0
down vote
Warning. This is a very wild solution for the first part of the problem (i.e., solving the homogeneous differential equation), intended only to illustrate its beauty (if you would agree), and not intended to be practical. This solution hinges on some heavy guess work, but it does not assume to know any solution to the required differential equaiton. That is, I was looking for a solution $g$ to the differential equation
$$g'(x)+big(g(x)big)^2-u(x),g(x)+v(x)=0,,tag$star$$$
where $u(x):=-dfrac2(2x-1)x(4x-1)$ and $v(x):=-dfrac4x(4x-1)$. I guessed that $g(x)=dfracax+bx(4x-1)$ for some $a$ and $b$, and then I solved for $a$ and $b$. I shall omit the gruesome detail of how I obtained $a$ and $b$, but it is really nothing but expanding terms in ($star$).
Let
$$f(x):=-frac8x-1x(4x-1)text and g(x):=frac4x+1x(4x-1),.$$ Observe that
$$f(x)+g(x)=u(x)text and f(x),g(x)-g'(x)=v(x),.$$
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2-u(X),D+v(X)=big(D-f(X)big),big(D-g(X)big),.tag*$$
Suppose that a function $y$ satisfies $$big(D^2-u(X),D+v(X)big),y=0,.tag#$$ Let $z:=big(D-g(X)big),y$. By (*) and (#),
$$big(D-f(X)big),z=0,.$$
Consequently,
$$z'(x)-f(x),z(x)=0text or z(x)=expleft(int,f(x),textdxright),.$$
This is a not-so-difficult work, and the result is
$$z(x)=frackx(4x-1)text for some constant k,.$$
Now, we want to solve $big(D-g(X)big),y=z$, or $$y'(x)-g(x),y(x)=z(x),.$$ That is, $$y(x)=frac1mu(x),int,mu(x),z(x),textdx,,text with mu(x):=expleft(-int,g(x),textdxright),.$$ We see that $mu(x)=dfracx(4x-1)^2$. Therefore, $$y(x)=frac(4x-1)^2x,int,fracx(4x-1)^2,left(frackx(4x-1)right),textdx,.$$ Ergo, $$y(x)=frac(4x-1)^2x,Biggl(A+B,left(frac1(4x-1)^2right)Biggr),,$$ where $A$ is a constant and $B:=-dfrack8$. This means $$y(x)=A,left(frac(4x-1)^2xright)+Bleft(frac1xright),.$$
Thus, the given solution $y_1(x)=dfrac1x$ corresponds to $(A,B):=(0,1)$. The linear solution $y_2(x)=2x-1$ found by Isham comes from $(A,B):=left(dfrac18,-dfrac18right)$.
P.S. This solution is inspired by a method of solving the quantum harmonic oscillator problem. One such method writes $$D^2-X^2+I=(D-X),(D+X),,$$
where $I$ is the identity operator.
answered Jul 31 at 18:32
Batominovski
22.8k22776
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Warning. This is a very wild solution for the first part of the problem (i.e., solving the homogeneous differential equation), intended only to illustrate its beauty (if you would agree), and not intended to be practical. This solution hinges on some heavy guess work, but it does not assume to know any solution to the required differential equaiton. That is, I was looking for a solution $g$ to the differential equation
$$g'(x)+big(g(x)big)^2-u(x),g(x)+v(x)=0,,tag$star$$$
where $u(x):=-dfrac2(2x-1)x(4x-1)$ and $v(x):=-dfrac4x(4x-1)$. I guessed that $g(x)=dfracax+bx(4x-1)$ for some $a$ and $b$, and then I solved for $a$ and $b$. I shall omit the gruesome detail of how I obtained $a$ and $b$, but it is really nothing but expanding terms in ($star$).
Let
$$f(x):=-frac8x-1x(4x-1)text and g(x):=frac4x+1x(4x-1),.$$ Observe that
$$f(x)+g(x)=u(x)text and f(x),g(x)-g'(x)=v(x),.$$
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2-u(X),D+v(X)=big(D-f(X)big),big(D-g(X)big),.tag*$$
Suppose that a function $y$ satisfies $$big(D^2-u(X),D+v(X)big),y=0,.tag#$$ Let $z:=big(D-g(X)big),y$. By (*) and (#),
$$big(D-f(X)big),z=0,.$$
Consequently,
$$z'(x)-f(x),z(x)=0text or z(x)=expleft(int,f(x),textdxright),.$$
This is a not-so-difficult work, and the result is
$$z(x)=frackx(4x-1)text for some constant k,.$$
Now, we want to solve $big(D-g(X)big),y=z$, or $$y'(x)-g(x),y(x)=z(x),.$$ That is, $$y(x)=frac1mu(x),int,mu(x),z(x),textdx,,text with mu(x):=expleft(-int,g(x),textdxright),.$$ We see that $mu(x)=dfracx(4x-1)^2$. Therefore, $$y(x)=frac(4x-1)^2x,int,fracx(4x-1)^2,left(frackx(4x-1)right),textdx,.$$ Ergo, $$y(x)=frac(4x-1)^2x,Biggl(A+B,left(frac1(4x-1)^2right)Biggr),,$$ where $A$ is a constant and $B:=-dfrack8$. This means $$y(x)=A,left(frac(4x-1)^2xright)+Bleft(frac1xright),.$$
Thus, the given solution $y_1(x)=dfrac1x$ corresponds to $(A,B):=(0,1)$. The linear solution $y_2(x)=2x-1$ found by Isham comes from $(A,B):=left(dfrac18,-dfrac18right)$.
P.S. This solution is inspired by a method of solving the quantum harmonic oscillator problem. One such method writes $$D^2-X^2+I=(D-X),(D+X),,$$
where $I$ is the identity operator.
answered Jul 31 at 18:32
Batominovski
22.8k22776
Warning. This is a very wild solution for the first part of the problem (i.e., solving the homogeneous differential equation), intended only to illustrate its beauty (if you would agree), and not intended to be practical. This solution hinges on some heavy guess work, but it does not assume to know any solution to the required differential equaiton. That is, I was looking for a solution $g$ to the differential equation
$$g'(x)+big(g(x)big)^2-u(x),g(x)+v(x)=0,,tag$star$$$
where $u(x):=-dfrac2(2x-1)x(4x-1)$ and $v(x):=-dfrac4x(4x-1)$. I guessed that $g(x)=dfracax+bx(4x-1)$ for some $a$ and $b$, and then I solved for $a$ and $b$. I shall omit the gruesome detail of how I obtained $a$ and $b$, but it is really nothing but expanding terms in ($star$).
Let
$$f(x):=-frac8x-1x(4x-1)text and g(x):=frac4x+1x(4x-1),.$$ Observe that
$$f(x)+g(x)=u(x)text and f(x),g(x)-g'(x)=v(x),.$$
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2-u(X),D+v(X)=big(D-f(X)big),big(D-g(X)big),.tag*$$
Suppose that a function $y$ satisfies $$big(D^2-u(X),D+v(X)big),y=0,.tag#$$ Let $z:=big(D-g(X)big),y$. By (*) and (#),
$$big(D-f(X)big),z=0,.$$
Consequently,
$$z'(x)-f(x),z(x)=0text or z(x)=expleft(int,f(x),textdxright),.$$
This is a not-so-difficult work, and the result is
$$z(x)=frackx(4x-1)text for some constant k,.$$
Now, we want to solve $big(D-g(X)big),y=z$, or $$y'(x)-g(x),y(x)=z(x),.$$ That is, $$y(x)=frac1mu(x),int,mu(x),z(x),textdx,,text with mu(x):=expleft(-int,g(x),textdxright),.$$ We see that $mu(x)=dfracx(4x-1)^2$. Therefore, $$y(x)=frac(4x-1)^2x,int,fracx(4x-1)^2,left(frackx(4x-1)right),textdx,.$$ Ergo, $$y(x)=frac(4x-1)^2x,Biggl(A+B,left(frac1(4x-1)^2right)Biggr),,$$ where $A$ is a constant and $B:=-dfrack8$. This means $$y(x)=A,left(frac(4x-1)^2xright)+Bleft(frac1xright),.$$
Thus, the given solution $y_1(x)=dfrac1x$ corresponds to $(A,B):=(0,1)$. The linear solution $y_2(x)=2x-1$ found by Isham comes from $(A,B):=left(dfrac18,-dfrac18right)$.
P.S. This solution is inspired by a method of solving the quantum harmonic oscillator problem. One such method writes $$D^2-X^2+I=(D-X),(D+X),,$$
where $I$ is the identity operator.
answered Jul 31 at 18:32
Batominovski
22.8k22776
edited Jul 31 at 20:01
answered Jul 31 at 18:32
Batominovski
22.8k22776
answered Jul 31 at 18:32
Batominovski
22.8k22776
answered Jul 31 at 18:32
Batominovski
22.8k22776
22.8k22776
add a comment |Â
add a comment |Â
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what is $y1$ again ?
– Ahmad Bazzi
Jul 31 at 15:15
1
@Liz: Are you sure you wrote everything correctly?
– Moo
Jul 31 at 15:24
Thanks .. should have checked before posting :)
– Liz
Jul 31 at 15:29