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I am self studying differential equations, and ran into some problems I have never seen before.
The majority of problems I have solved have come int the form of $Ay''-by'+C = 0$. I clearly understand how to solve these, and arrive at the general solutions.
However, I am now encountering a problem like
$2t^2y''-5ty'+6y = 0$ where they give a solution $y(t) = t^frac32)$.
Does this now mean I have to work backwards from before since we already have a solution given to us?
Likewise, how does the general form look different in the case when
$2t^2y'' - 5ty' + 6t = t^2 - 1$. I know this revolves around methods of undetermined coefficients, and have guessed $Ax^2+Bx+C$ and have differentiating it twice over for y'' and once over for y', but I still haven't organized my work together to arrive at a general solution.
Thank you
differential-equations nonlinear-system
asked yesterday
DoofusAnarchy
74
add a comment |Â
up vote
0
down vote
favorite
I am self studying differential equations, and ran into some problems I have never seen before.
The majority of problems I have solved have come int the form of $Ay''-by'+C = 0$. I clearly understand how to solve these, and arrive at the general solutions.
However, I am now encountering a problem like
$2t^2y''-5ty'+6y = 0$ where they give a solution $y(t) = t^frac32)$.
Does this now mean I have to work backwards from before since we already have a solution given to us?
Likewise, how does the general form look different in the case when
$2t^2y'' - 5ty' + 6t = t^2 - 1$. I know this revolves around methods of undetermined coefficients, and have guessed $Ax^2+Bx+C$ and have differentiating it twice over for y'' and once over for y', but I still haven't organized my work together to arrive at a general solution.
Thank you
differential-equations nonlinear-system
asked yesterday
DoofusAnarchy
74
en.wikipedia.org/wiki/Frobenius_method
– Lord Shark the Unknown
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am self studying differential equations, and ran into some problems I have never seen before.
The majority of problems I have solved have come int the form of $Ay''-by'+C = 0$. I clearly understand how to solve these, and arrive at the general solutions.
However, I am now encountering a problem like
$2t^2y''-5ty'+6y = 0$ where they give a solution $y(t) = t^frac32)$.
Does this now mean I have to work backwards from before since we already have a solution given to us?
Likewise, how does the general form look different in the case when
$2t^2y'' - 5ty' + 6t = t^2 - 1$. I know this revolves around methods of undetermined coefficients, and have guessed $Ax^2+Bx+C$ and have differentiating it twice over for y'' and once over for y', but I still haven't organized my work together to arrive at a general solution.
Thank you
differential-equations nonlinear-system
asked yesterday
DoofusAnarchy
74
I am self studying differential equations, and ran into some problems I have never seen before.
The majority of problems I have solved have come int the form of $Ay''-by'+C = 0$. I clearly understand how to solve these, and arrive at the general solutions.
However, I am now encountering a problem like
$2t^2y''-5ty'+6y = 0$ where they give a solution $y(t) = t^frac32)$.
Does this now mean I have to work backwards from before since we already have a solution given to us?
Likewise, how does the general form look different in the case when
$2t^2y'' - 5ty' + 6t = t^2 - 1$. I know this revolves around methods of undetermined coefficients, and have guessed $Ax^2+Bx+C$ and have differentiating it twice over for y'' and once over for y', but I still haven't organized my work together to arrive at a general solution.
Thank you
differential-equations nonlinear-system
asked yesterday
DoofusAnarchy
74
edited yesterday
asked yesterday
DoofusAnarchy
74
asked yesterday
DoofusAnarchy
74
asked yesterday
DoofusAnarchy
74
74
en.wikipedia.org/wiki/Frobenius_method
– Lord Shark the Unknown
yesterday
add a comment |Â
en.wikipedia.org/wiki/Frobenius_method
– Lord Shark the Unknown
yesterday
en.wikipedia.org/wiki/Frobenius_method
– Lord Shark the Unknown
yesterday
en.wikipedia.org/wiki/Frobenius_method
– Lord Shark the Unknown
yesterday
add a comment |Â
3 Answers
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Hint: The differential equation
$$2t^2y''-5ty'+6y = 0$$
or
$$4t^2y''-10ty'+12y = 0$$
known as Euler differential equation, which can be solved with substitution $x=ln2t$. In this case when the right side is zero, the differential equation
$$(at)^2y''+paty'+qy = 0$$
is homogeneous and with $x=ln at$ is converted to
$$y''+left(dfracpa-1right)y'+dfracpa^2y = 0$$
then with characteristic equation
$$lambda^2+left(dfracpa-1right)lambda+dfracpa^2=0$$
we may find it's general solution.
Edit:
After substitution $x=ln 2t$, your equation is
$$y''-dfrac72y'+3y=2e^2x-2tag*$$
from characteristic equation, the general solution is
$$y_g=C_1e^2x+C_2e^frac32x$$
and while the right side of equation (*) has item $e^2x$ which is one of the general solutions, then let $y=Axe^2x+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=ln 2t$ and return main independent variable. One may find further examples here.
edited yesterday
mathreadler
13.5k71757
answered yesterday
user 108128
18.6k41544
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
 |Â
show 11 more comments
up vote
1
down vote
I think that it's informative to look at degrees, considering $t$ as raising the degree with one and $fracddt$ as lowering the degree with one. For all the terms in your differential equation you have as many $t$ factors as applications of $fracddt$ and the degree therefore remains unchanged. Even more important is that the degree changes in the same way for every term. This means that if the solution in some way can be written as a sum $y(t) = sum_alpha in A c_alpha t^alpha$ then the differential equation works on each term separately. In this case you get $2t^2 y'' - 5ty' + 6y = sum_alpha in A (2alpha(alpha-1)-5alpha+6) c_alpha t^alpha.$ so for the differential equation to be satisfied you must have $(2alpha(alpha-1)-5alpha+6) c_alpha = 0,$ i.e. $2(alpha-2)(alpha-frac32)c_alpha=0.$ Thus $c_alpha$ can only be nonzero for $alpha=2$ and $alpha=frac32.$
answered yesterday
md2perpe
5,5691821
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This is an alternative route, but I think it is better to follow the answer by user 108128. Rewrite the differential equation as
$$y''(t)-frac52t,y'(t)+frac3t^2,y(t)=f(t),,$$
where $f$ is a source function. (Note that $fequiv 0$ in the homogeneous case, and $$f(t)=fract^2-12t^2=frac12left(1-frac1t^2right)$$ for the nonhomogeneous equation in question.) Observe that $$fractextdtextdt,left(y'(t)-frac2t,y(t)right)-frac12t,left(y'(t)-frac2t,y(t)right)=f(t),.$$
Consequently,
$$fractextdtextdt,frac1sqrtt,left(y'(t)-frac2t,y(t)right)=frac1sqrtt,f(t),.$$
Ergo,
$$y'(t)-frac2t,y(t)=sqrtt,F(t),,$$
where $displaystyle F(t):=int,frac1sqrtt,f(t),textdt$. Finally, we write
$$fractextdtextdt,left(frac1t^2,y(t)right)=t^-frac32,F(t),,$$
so that
$$y(t)=t^2,int,t^-frac32,F(t),textdt,.$$
To solve the homogeneous solution (i.e., with $fequiv 0$, we have $F(t)=-frac12,A$ for some constant $A$. That is, $$y(t)=t^2,int,left(-frac12,Aright),t^-frac32,textdt=t^2,left(A,t^-frac12+Bright)=A,t^frac32+B,t^2,,$$ for some constant $B$.
For the nonhomogeneous solution with $f(t)=displaystylefrac12,left(1-frac1t^2right)$, we get $$F(t)=int,frac12left(t^-frac12-t^-frac52right),textdt=t^frac12+frac13,t^-frac32-frac12,a,,$$ where $a$ is a constant. That is, $$y(t)=t^2,int,t^-frac32,left(t^frac12+frac13,t^-frac32-frac12,aright),textdt=t^2,left(ln(t)-frac16t^-2+a,t^-frac12+bright),,$$ for some constant $b$. Thus, $$y(t)=t^2,ln(t)-frac16+a,t^frac32+b,t^2,.$$
P.S. You can also rewrite the differential equation as
$$fractextdtextdt,left(y'(t)-frac32t,y(t)right)-frac1t,left(y'(t)-frac32t,y(t)right)=f(t),.$$
This leads to
$$y(t)=t^frac32,int,t^-frac12,tildeF(t),textdt,,$$
where $$tildeF(t):=int,frac1t,f(t),textdt,.$$
answered yesterday
Batominovski
22.2k22676
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
 |Â
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: The differential equation
$$2t^2y''-5ty'+6y = 0$$
or
$$4t^2y''-10ty'+12y = 0$$
known as Euler differential equation, which can be solved with substitution $x=ln2t$. In this case when the right side is zero, the differential equation
$$(at)^2y''+paty'+qy = 0$$
is homogeneous and with $x=ln at$ is converted to
$$y''+left(dfracpa-1right)y'+dfracpa^2y = 0$$
then with characteristic equation
$$lambda^2+left(dfracpa-1right)lambda+dfracpa^2=0$$
we may find it's general solution.
Edit:
After substitution $x=ln 2t$, your equation is
$$y''-dfrac72y'+3y=2e^2x-2tag*$$
from characteristic equation, the general solution is
$$y_g=C_1e^2x+C_2e^frac32x$$
and while the right side of equation (*) has item $e^2x$ which is one of the general solutions, then let $y=Axe^2x+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=ln 2t$ and return main independent variable. One may find further examples here.
edited yesterday
mathreadler
13.5k71757
answered yesterday
user 108128
18.6k41544
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
 |Â
show 11 more comments
up vote
1
down vote
accepted
Hint: The differential equation
$$2t^2y''-5ty'+6y = 0$$
or
$$4t^2y''-10ty'+12y = 0$$
known as Euler differential equation, which can be solved with substitution $x=ln2t$. In this case when the right side is zero, the differential equation
$$(at)^2y''+paty'+qy = 0$$
is homogeneous and with $x=ln at$ is converted to
$$y''+left(dfracpa-1right)y'+dfracpa^2y = 0$$
then with characteristic equation
$$lambda^2+left(dfracpa-1right)lambda+dfracpa^2=0$$
we may find it's general solution.
Edit:
After substitution $x=ln 2t$, your equation is
$$y''-dfrac72y'+3y=2e^2x-2tag*$$
from characteristic equation, the general solution is
$$y_g=C_1e^2x+C_2e^frac32x$$
and while the right side of equation (*) has item $e^2x$ which is one of the general solutions, then let $y=Axe^2x+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=ln 2t$ and return main independent variable. One may find further examples here.
edited yesterday
mathreadler
13.5k71757
answered yesterday
user 108128
18.6k41544
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
 |Â
show 11 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: The differential equation
$$2t^2y''-5ty'+6y = 0$$
or
$$4t^2y''-10ty'+12y = 0$$
known as Euler differential equation, which can be solved with substitution $x=ln2t$. In this case when the right side is zero, the differential equation
$$(at)^2y''+paty'+qy = 0$$
is homogeneous and with $x=ln at$ is converted to
$$y''+left(dfracpa-1right)y'+dfracpa^2y = 0$$
then with characteristic equation
$$lambda^2+left(dfracpa-1right)lambda+dfracpa^2=0$$
we may find it's general solution.
Edit:
After substitution $x=ln 2t$, your equation is
$$y''-dfrac72y'+3y=2e^2x-2tag*$$
from characteristic equation, the general solution is
$$y_g=C_1e^2x+C_2e^frac32x$$
and while the right side of equation (*) has item $e^2x$ which is one of the general solutions, then let $y=Axe^2x+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=ln 2t$ and return main independent variable. One may find further examples here.
edited yesterday
mathreadler
13.5k71757
answered yesterday
user 108128
18.6k41544
Hint: The differential equation
$$2t^2y''-5ty'+6y = 0$$
or
$$4t^2y''-10ty'+12y = 0$$
known as Euler differential equation, which can be solved with substitution $x=ln2t$. In this case when the right side is zero, the differential equation
$$(at)^2y''+paty'+qy = 0$$
is homogeneous and with $x=ln at$ is converted to
$$y''+left(dfracpa-1right)y'+dfracpa^2y = 0$$
then with characteristic equation
$$lambda^2+left(dfracpa-1right)lambda+dfracpa^2=0$$
we may find it's general solution.
Edit:
After substitution $x=ln 2t$, your equation is
$$y''-dfrac72y'+3y=2e^2x-2tag*$$
from characteristic equation, the general solution is
$$y_g=C_1e^2x+C_2e^frac32x$$
and while the right side of equation (*) has item $e^2x$ which is one of the general solutions, then let $y=Axe^2x+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=ln 2t$ and return main independent variable. One may find further examples here.
edited yesterday
mathreadler
13.5k71757
answered yesterday
user 108128
18.6k41544
edited yesterday
mathreadler
13.5k71757
edited yesterday
mathreadler
13.5k71757
edited yesterday
mathreadler
13.5k71757
13.5k71757
answered yesterday
user 108128
18.6k41544
answered yesterday
user 108128
18.6k41544
answered yesterday
user 108128
18.6k41544
18.6k41544
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
 |Â
show 11 more comments
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I have researched online for my question to Part A, and it suggests that I can use the Wronskian to derive another fundamental solution, which I can work backwards to find another fundamental solution and attain the general solution? Would that method work. Also, is my guess correct regarding my second equation?
– DoofusAnarchy
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
I've added some details. Can you proceed?
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
If you have a solution, you can fin the second with $$y_2=y_1intfrace^-int p,dxy_1^2$$
– user 108128
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
Thank you for all this information. Where does the condition that y = t^3/2 come into play in this problem
– DoofusAnarchy
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
You are welcome and try more with examples, you will find the problem easy!
– user 108128
yesterday
 |Â
show 11 more comments
up vote
1
down vote
I think that it's informative to look at degrees, considering $t$ as raising the degree with one and $fracddt$ as lowering the degree with one. For all the terms in your differential equation you have as many $t$ factors as applications of $fracddt$ and the degree therefore remains unchanged. Even more important is that the degree changes in the same way for every term. This means that if the solution in some way can be written as a sum $y(t) = sum_alpha in A c_alpha t^alpha$ then the differential equation works on each term separately. In this case you get $2t^2 y'' - 5ty' + 6y = sum_alpha in A (2alpha(alpha-1)-5alpha+6) c_alpha t^alpha.$ so for the differential equation to be satisfied you must have $(2alpha(alpha-1)-5alpha+6) c_alpha = 0,$ i.e. $2(alpha-2)(alpha-frac32)c_alpha=0.$ Thus $c_alpha$ can only be nonzero for $alpha=2$ and $alpha=frac32.$
answered yesterday
md2perpe
5,5691821
add a comment |Â
up vote
1
down vote
I think that it's informative to look at degrees, considering $t$ as raising the degree with one and $fracddt$ as lowering the degree with one. For all the terms in your differential equation you have as many $t$ factors as applications of $fracddt$ and the degree therefore remains unchanged. Even more important is that the degree changes in the same way for every term. This means that if the solution in some way can be written as a sum $y(t) = sum_alpha in A c_alpha t^alpha$ then the differential equation works on each term separately. In this case you get $2t^2 y'' - 5ty' + 6y = sum_alpha in A (2alpha(alpha-1)-5alpha+6) c_alpha t^alpha.$ so for the differential equation to be satisfied you must have $(2alpha(alpha-1)-5alpha+6) c_alpha = 0,$ i.e. $2(alpha-2)(alpha-frac32)c_alpha=0.$ Thus $c_alpha$ can only be nonzero for $alpha=2$ and $alpha=frac32.$
answered yesterday
md2perpe
5,5691821
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think that it's informative to look at degrees, considering $t$ as raising the degree with one and $fracddt$ as lowering the degree with one. For all the terms in your differential equation you have as many $t$ factors as applications of $fracddt$ and the degree therefore remains unchanged. Even more important is that the degree changes in the same way for every term. This means that if the solution in some way can be written as a sum $y(t) = sum_alpha in A c_alpha t^alpha$ then the differential equation works on each term separately. In this case you get $2t^2 y'' - 5ty' + 6y = sum_alpha in A (2alpha(alpha-1)-5alpha+6) c_alpha t^alpha.$ so for the differential equation to be satisfied you must have $(2alpha(alpha-1)-5alpha+6) c_alpha = 0,$ i.e. $2(alpha-2)(alpha-frac32)c_alpha=0.$ Thus $c_alpha$ can only be nonzero for $alpha=2$ and $alpha=frac32.$
answered yesterday
md2perpe
5,5691821
I think that it's informative to look at degrees, considering $t$ as raising the degree with one and $fracddt$ as lowering the degree with one. For all the terms in your differential equation you have as many $t$ factors as applications of $fracddt$ and the degree therefore remains unchanged. Even more important is that the degree changes in the same way for every term. This means that if the solution in some way can be written as a sum $y(t) = sum_alpha in A c_alpha t^alpha$ then the differential equation works on each term separately. In this case you get $2t^2 y'' - 5ty' + 6y = sum_alpha in A (2alpha(alpha-1)-5alpha+6) c_alpha t^alpha.$ so for the differential equation to be satisfied you must have $(2alpha(alpha-1)-5alpha+6) c_alpha = 0,$ i.e. $2(alpha-2)(alpha-frac32)c_alpha=0.$ Thus $c_alpha$ can only be nonzero for $alpha=2$ and $alpha=frac32.$
answered yesterday
md2perpe
5,5691821
answered yesterday
md2perpe
5,5691821
answered yesterday
md2perpe
5,5691821
answered yesterday
md2perpe
5,5691821
5,5691821
add a comment |Â
add a comment |Â
up vote
0
down vote
This is an alternative route, but I think it is better to follow the answer by user 108128. Rewrite the differential equation as
$$y''(t)-frac52t,y'(t)+frac3t^2,y(t)=f(t),,$$
where $f$ is a source function. (Note that $fequiv 0$ in the homogeneous case, and $$f(t)=fract^2-12t^2=frac12left(1-frac1t^2right)$$ for the nonhomogeneous equation in question.) Observe that $$fractextdtextdt,left(y'(t)-frac2t,y(t)right)-frac12t,left(y'(t)-frac2t,y(t)right)=f(t),.$$
Consequently,
$$fractextdtextdt,frac1sqrtt,left(y'(t)-frac2t,y(t)right)=frac1sqrtt,f(t),.$$
Ergo,
$$y'(t)-frac2t,y(t)=sqrtt,F(t),,$$
where $displaystyle F(t):=int,frac1sqrtt,f(t),textdt$. Finally, we write
$$fractextdtextdt,left(frac1t^2,y(t)right)=t^-frac32,F(t),,$$
so that
$$y(t)=t^2,int,t^-frac32,F(t),textdt,.$$
To solve the homogeneous solution (i.e., with $fequiv 0$, we have $F(t)=-frac12,A$ for some constant $A$. That is, $$y(t)=t^2,int,left(-frac12,Aright),t^-frac32,textdt=t^2,left(A,t^-frac12+Bright)=A,t^frac32+B,t^2,,$$ for some constant $B$.
For the nonhomogeneous solution with $f(t)=displaystylefrac12,left(1-frac1t^2right)$, we get $$F(t)=int,frac12left(t^-frac12-t^-frac52right),textdt=t^frac12+frac13,t^-frac32-frac12,a,,$$ where $a$ is a constant. That is, $$y(t)=t^2,int,t^-frac32,left(t^frac12+frac13,t^-frac32-frac12,aright),textdt=t^2,left(ln(t)-frac16t^-2+a,t^-frac12+bright),,$$ for some constant $b$. Thus, $$y(t)=t^2,ln(t)-frac16+a,t^frac32+b,t^2,.$$
P.S. You can also rewrite the differential equation as
$$fractextdtextdt,left(y'(t)-frac32t,y(t)right)-frac1t,left(y'(t)-frac32t,y(t)right)=f(t),.$$
This leads to
$$y(t)=t^frac32,int,t^-frac12,tildeF(t),textdt,,$$
where $$tildeF(t):=int,frac1t,f(t),textdt,.$$
answered yesterday
Batominovski
22.2k22676
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
 |Â
show 10 more comments
up vote
0
down vote
This is an alternative route, but I think it is better to follow the answer by user 108128. Rewrite the differential equation as
$$y''(t)-frac52t,y'(t)+frac3t^2,y(t)=f(t),,$$
where $f$ is a source function. (Note that $fequiv 0$ in the homogeneous case, and $$f(t)=fract^2-12t^2=frac12left(1-frac1t^2right)$$ for the nonhomogeneous equation in question.) Observe that $$fractextdtextdt,left(y'(t)-frac2t,y(t)right)-frac12t,left(y'(t)-frac2t,y(t)right)=f(t),.$$
Consequently,
$$fractextdtextdt,frac1sqrtt,left(y'(t)-frac2t,y(t)right)=frac1sqrtt,f(t),.$$
Ergo,
$$y'(t)-frac2t,y(t)=sqrtt,F(t),,$$
where $displaystyle F(t):=int,frac1sqrtt,f(t),textdt$. Finally, we write
$$fractextdtextdt,left(frac1t^2,y(t)right)=t^-frac32,F(t),,$$
so that
$$y(t)=t^2,int,t^-frac32,F(t),textdt,.$$
To solve the homogeneous solution (i.e., with $fequiv 0$, we have $F(t)=-frac12,A$ for some constant $A$. That is, $$y(t)=t^2,int,left(-frac12,Aright),t^-frac32,textdt=t^2,left(A,t^-frac12+Bright)=A,t^frac32+B,t^2,,$$ for some constant $B$.
For the nonhomogeneous solution with $f(t)=displaystylefrac12,left(1-frac1t^2right)$, we get $$F(t)=int,frac12left(t^-frac12-t^-frac52right),textdt=t^frac12+frac13,t^-frac32-frac12,a,,$$ where $a$ is a constant. That is, $$y(t)=t^2,int,t^-frac32,left(t^frac12+frac13,t^-frac32-frac12,aright),textdt=t^2,left(ln(t)-frac16t^-2+a,t^-frac12+bright),,$$ for some constant $b$. Thus, $$y(t)=t^2,ln(t)-frac16+a,t^frac32+b,t^2,.$$
P.S. You can also rewrite the differential equation as
$$fractextdtextdt,left(y'(t)-frac32t,y(t)right)-frac1t,left(y'(t)-frac32t,y(t)right)=f(t),.$$
This leads to
$$y(t)=t^frac32,int,t^-frac12,tildeF(t),textdt,,$$
where $$tildeF(t):=int,frac1t,f(t),textdt,.$$
answered yesterday
Batominovski
22.2k22676
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
 |Â
show 10 more comments
up vote
0
down vote
up vote
0
down vote
This is an alternative route, but I think it is better to follow the answer by user 108128. Rewrite the differential equation as
$$y''(t)-frac52t,y'(t)+frac3t^2,y(t)=f(t),,$$
where $f$ is a source function. (Note that $fequiv 0$ in the homogeneous case, and $$f(t)=fract^2-12t^2=frac12left(1-frac1t^2right)$$ for the nonhomogeneous equation in question.) Observe that $$fractextdtextdt,left(y'(t)-frac2t,y(t)right)-frac12t,left(y'(t)-frac2t,y(t)right)=f(t),.$$
Consequently,
$$fractextdtextdt,frac1sqrtt,left(y'(t)-frac2t,y(t)right)=frac1sqrtt,f(t),.$$
Ergo,
$$y'(t)-frac2t,y(t)=sqrtt,F(t),,$$
where $displaystyle F(t):=int,frac1sqrtt,f(t),textdt$. Finally, we write
$$fractextdtextdt,left(frac1t^2,y(t)right)=t^-frac32,F(t),,$$
so that
$$y(t)=t^2,int,t^-frac32,F(t),textdt,.$$
To solve the homogeneous solution (i.e., with $fequiv 0$, we have $F(t)=-frac12,A$ for some constant $A$. That is, $$y(t)=t^2,int,left(-frac12,Aright),t^-frac32,textdt=t^2,left(A,t^-frac12+Bright)=A,t^frac32+B,t^2,,$$ for some constant $B$.
For the nonhomogeneous solution with $f(t)=displaystylefrac12,left(1-frac1t^2right)$, we get $$F(t)=int,frac12left(t^-frac12-t^-frac52right),textdt=t^frac12+frac13,t^-frac32-frac12,a,,$$ where $a$ is a constant. That is, $$y(t)=t^2,int,t^-frac32,left(t^frac12+frac13,t^-frac32-frac12,aright),textdt=t^2,left(ln(t)-frac16t^-2+a,t^-frac12+bright),,$$ for some constant $b$. Thus, $$y(t)=t^2,ln(t)-frac16+a,t^frac32+b,t^2,.$$
P.S. You can also rewrite the differential equation as
$$fractextdtextdt,left(y'(t)-frac32t,y(t)right)-frac1t,left(y'(t)-frac32t,y(t)right)=f(t),.$$
This leads to
$$y(t)=t^frac32,int,t^-frac12,tildeF(t),textdt,,$$
where $$tildeF(t):=int,frac1t,f(t),textdt,.$$
answered yesterday
Batominovski
22.2k22676
This is an alternative route, but I think it is better to follow the answer by user 108128. Rewrite the differential equation as
$$y''(t)-frac52t,y'(t)+frac3t^2,y(t)=f(t),,$$
where $f$ is a source function. (Note that $fequiv 0$ in the homogeneous case, and $$f(t)=fract^2-12t^2=frac12left(1-frac1t^2right)$$ for the nonhomogeneous equation in question.) Observe that $$fractextdtextdt,left(y'(t)-frac2t,y(t)right)-frac12t,left(y'(t)-frac2t,y(t)right)=f(t),.$$
Consequently,
$$fractextdtextdt,frac1sqrtt,left(y'(t)-frac2t,y(t)right)=frac1sqrtt,f(t),.$$
Ergo,
$$y'(t)-frac2t,y(t)=sqrtt,F(t),,$$
where $displaystyle F(t):=int,frac1sqrtt,f(t),textdt$. Finally, we write
$$fractextdtextdt,left(frac1t^2,y(t)right)=t^-frac32,F(t),,$$
so that
$$y(t)=t^2,int,t^-frac32,F(t),textdt,.$$
To solve the homogeneous solution (i.e., with $fequiv 0$, we have $F(t)=-frac12,A$ for some constant $A$. That is, $$y(t)=t^2,int,left(-frac12,Aright),t^-frac32,textdt=t^2,left(A,t^-frac12+Bright)=A,t^frac32+B,t^2,,$$ for some constant $B$.
For the nonhomogeneous solution with $f(t)=displaystylefrac12,left(1-frac1t^2right)$, we get $$F(t)=int,frac12left(t^-frac12-t^-frac52right),textdt=t^frac12+frac13,t^-frac32-frac12,a,,$$ where $a$ is a constant. That is, $$y(t)=t^2,int,t^-frac32,left(t^frac12+frac13,t^-frac32-frac12,aright),textdt=t^2,left(ln(t)-frac16t^-2+a,t^-frac12+bright),,$$ for some constant $b$. Thus, $$y(t)=t^2,ln(t)-frac16+a,t^frac32+b,t^2,.$$
P.S. You can also rewrite the differential equation as
$$fractextdtextdt,left(y'(t)-frac32t,y(t)right)-frac1t,left(y'(t)-frac32t,y(t)right)=f(t),.$$
This leads to
$$y(t)=t^frac32,int,t^-frac12,tildeF(t),textdt,,$$
where $$tildeF(t):=int,frac1t,f(t),textdt,.$$
answered yesterday
Batominovski
22.2k22676
edited yesterday
answered yesterday
Batominovski
22.2k22676
answered yesterday
Batominovski
22.2k22676
answered yesterday
Batominovski
22.2k22676
22.2k22676
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
 |Â
show 10 more comments
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
I used Ax^2 + Bx+ C for my guess for the nonhomogenous equation. I attained A = 1/10...B = -1/6 ...and C = -1/36.... Why does my solution vary drastically from yours
– DoofusAnarchy
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
That doesn't work. If you plug in $y(t)=At^2+Bt+C$ into the nonhomogeneous equation, you will get $$Bt+6C=t^2-1,.$$ But a linear polynomial is never equal to a quadratic polynomial.
– Batominovski
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
hmmm but I always assumed you make your guess based on the right side. You at least, and at most, want to match the highest polynomial. Why doesn't a standard polynomial guess not work?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
say for example I have the equation y''-2y+2y = 2e^t. I would base my guess on the RHS. So wouldn't my guess just simply be Ae^t for this particular question?
– DoofusAnarchy
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
It is actually the other way around. If a guess works, then it is mostly a miracle (usually because the problem is designed to allow certain forms of educated guesses). But if a random problem is thrown at you, chances are that none of your known elementary functions will work. There is no particular reason why certain guesses work, why some do not.
– Batominovski
yesterday
 |Â
show 10 more comments
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en.wikipedia.org/wiki/Frobenius_method
– Lord Shark the Unknown
yesterday