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Reduction of Order Leads to Non-Elementary Integral
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If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
beginalign
fracdwdx(x^2+x)-w(x^2+1)&=0 \
fracdwdx&=frac(x^2+x)^-1(x^2+1)w^-1 \
textln(w)&=int fracx^2+1x^2+x dx \
textln(w)&=int 1-frac1x+frac2x+1 dx \
textln(w)&=x+textlnleft(frac(x+1)^2xright)+C \
w&=C_1frace^x(x+1)^2x \
v&=C_1int frace^x(x+1)^2x dx
endalign
Where $$C_1int frace^xx dx$$ cannot be solved. How do I find $v$?
calculus integration differential-equations reduction-of-order-ode
asked Jul 29 at 11:38
Bell
565113
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show 1 more comment
up vote
2
down vote
favorite
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
beginalign
fracdwdx(x^2+x)-w(x^2+1)&=0 \
fracdwdx&=frac(x^2+x)^-1(x^2+1)w^-1 \
textln(w)&=int fracx^2+1x^2+x dx \
textln(w)&=int 1-frac1x+frac2x+1 dx \
textln(w)&=x+textlnleft(frac(x+1)^2xright)+C \
w&=C_1frace^x(x+1)^2x \
v&=C_1int frace^x(x+1)^2x dx
endalign
Where $$C_1int frace^xx dx$$ cannot be solved. How do I find $v$?
calculus integration differential-equations reduction-of-order-ode
asked Jul 29 at 11:38
Bell
565113
4
There's nothing wrong with leaving the solution as an integral. Also, the last integral is the exponential integral
– Dylan
Jul 29 at 11:51
The answer provided is $u_2=Ae^x$. I'd like to know, if possible, how they got this answer
– Bell
Jul 29 at 11:54
1
If that's the case, then you must have done something wrong. I'd go back and check if your derived equation (second line) is correct
– Dylan
Jul 29 at 11:55
I will have another look, but I've read through it many times. I haven't found a mistake yet.
– Bell
Jul 29 at 11:58
4
Your partial fraction expansion has an error. The minus sign should be plus and the plus sign should be minus. Maybe that will fix the final answer.
– B. Goddard
Jul 29 at 12:08
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
beginalign
fracdwdx(x^2+x)-w(x^2+1)&=0 \
fracdwdx&=frac(x^2+x)^-1(x^2+1)w^-1 \
textln(w)&=int fracx^2+1x^2+x dx \
textln(w)&=int 1-frac1x+frac2x+1 dx \
textln(w)&=x+textlnleft(frac(x+1)^2xright)+C \
w&=C_1frace^x(x+1)^2x \
v&=C_1int frace^x(x+1)^2x dx
endalign
Where $$C_1int frace^xx dx$$ cannot be solved. How do I find $v$?
calculus integration differential-equations reduction-of-order-ode
asked Jul 29 at 11:38
Bell
565113
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
beginalign
fracdwdx(x^2+x)-w(x^2+1)&=0 \
fracdwdx&=frac(x^2+x)^-1(x^2+1)w^-1 \
textln(w)&=int fracx^2+1x^2+x dx \
textln(w)&=int 1-frac1x+frac2x+1 dx \
textln(w)&=x+textlnleft(frac(x+1)^2xright)+C \
w&=C_1frace^x(x+1)^2x \
v&=C_1int frace^x(x+1)^2x dx
endalign
Where $$C_1int frace^xx dx$$ cannot be solved. How do I find $v$?
calculus integration differential-equations reduction-of-order-ode
asked Jul 29 at 11:38
Bell
565113
asked Jul 29 at 11:38
Bell
565113
asked Jul 29 at 11:38
Bell
565113
asked Jul 29 at 11:38
Bell
565113
565113
4
There's nothing wrong with leaving the solution as an integral. Also, the last integral is the exponential integral
– Dylan
Jul 29 at 11:51
The answer provided is $u_2=Ae^x$. I'd like to know, if possible, how they got this answer
– Bell
Jul 29 at 11:54
1
If that's the case, then you must have done something wrong. I'd go back and check if your derived equation (second line) is correct
– Dylan
Jul 29 at 11:55
I will have another look, but I've read through it many times. I haven't found a mistake yet.
– Bell
Jul 29 at 11:58
4
Your partial fraction expansion has an error. The minus sign should be plus and the plus sign should be minus. Maybe that will fix the final answer.
– B. Goddard
Jul 29 at 12:08
 |Â
show 1 more comment
4
There's nothing wrong with leaving the solution as an integral. Also, the last integral is the exponential integral
– Dylan
Jul 29 at 11:51
The answer provided is $u_2=Ae^x$. I'd like to know, if possible, how they got this answer
– Bell
Jul 29 at 11:54
1
If that's the case, then you must have done something wrong. I'd go back and check if your derived equation (second line) is correct
– Dylan
Jul 29 at 11:55
I will have another look, but I've read through it many times. I haven't found a mistake yet.
– Bell
Jul 29 at 11:58
4
Your partial fraction expansion has an error. The minus sign should be plus and the plus sign should be minus. Maybe that will fix the final answer.
– B. Goddard
Jul 29 at 12:08
4
4
There's nothing wrong with leaving the solution as an integral. Also, the last integral is the exponential integral
– Dylan
Jul 29 at 11:51
There's nothing wrong with leaving the solution as an integral. Also, the last integral is the exponential integral
– Dylan
Jul 29 at 11:51
The answer provided is $u_2=Ae^x$. I'd like to know, if possible, how they got this answer
– Bell
Jul 29 at 11:54
The answer provided is $u_2=Ae^x$. I'd like to know, if possible, how they got this answer
– Bell
Jul 29 at 11:54
1
1
If that's the case, then you must have done something wrong. I'd go back and check if your derived equation (second line) is correct
– Dylan
Jul 29 at 11:55
If that's the case, then you must have done something wrong. I'd go back and check if your derived equation (second line) is correct
– Dylan
Jul 29 at 11:55
I will have another look, but I've read through it many times. I haven't found a mistake yet.
– Bell
Jul 29 at 11:58
I will have another look, but I've read through it many times. I haven't found a mistake yet.
– Bell
Jul 29 at 11:58
4
4
Your partial fraction expansion has an error. The minus sign should be plus and the plus sign should be minus. Maybe that will fix the final answer.
– B. Goddard
Jul 29 at 12:08
Your partial fraction expansion has an error. The minus sign should be plus and the plus sign should be minus. Maybe that will fix the final answer.
– B. Goddard
Jul 29 at 12:08
 |Â
show 1 more comment
3 Answers
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Your work is correct up to the partial fractions
$$ fracv''v' = fracx^2+1x^2+x = 1 + frac1-xx(x+1) = 1 + frac1x - frac2x+1 $$
Integrating this gives
$$ ln(v') = x + ln x - 2ln(x+1) $$
$$ implies v' = fracxe^x(x+1)^2 = frace^xx+1-frace^x(x+1)^2 $$
This has the form $e^xf(x) + e^xf'(x) = (e^xf(x))'$, therefore
$$ v(x) = frace^xx+1 $$
I've ignored the integration constants, since they're already included in the general solution, which is
$$ u(x) = (x+1)(c_1 + c_2v(x)) = c_1(x+1) + c_2e^x $$
answered Jul 29 at 12:20
Dylan
11.4k31026
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up vote
5
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$$xu''-(x+1)u'+u=0$$
$$x(u''-u')-(u'-u)=0$$
Let $v=u'-u$,
$$xv'-v=0$$
beginalign
xv' &= v \
fracdvdx &= fracvx \
ln v &= ln x + C \
v &= Ax \
u'-u &= Ax \
u &= dfrac1expleft(-int dxright)int Axexpleft(-int dxright) dx \
&= Ae^xint xe^-x dx \
&= Ae^x left[-xe^-x-int-e^-xdxright] \
&= Ae^x left[-(x+1)e^-x+Bright] \
&= k_1(x+1)+k_2e^x \
endalign
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
add a comment |Â
up vote
1
down vote
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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-left(fracX+1Xright),D+frac1X=left(D-1-frac1X+frac1X+1right),left(D-frac1X+1right),.$$
First, let $v:=left(D-dfrac1X+1right),u$. Then, $left(D-1-dfrac1X+dfrac1X+1right),v=0$. Therefore,
$$v(x)=expBiggl(int,left(1+frac1x-frac1x+1right),textdxBiggr)=a,left(fracx,exp(x)x+1right)$$
for some constant $a$. Now, $v=left(D-dfrac1X+1right),u$ implies that $D,left(dfrac1X+1,uright)=dfrac1X+1,v$. Hence,
$$u(x)=(x+1),int,fracx,exp(x)(x+1)^2,textdx=(x+1),Biggl(aleft(fracexp(x)x+1right)+bBiggr)=a,exp(x)+b(x+1),,$$
for some constant $b$.
Interestingly, you can also write
$$D^2-left(fracX+1Xright),D+frac1X=left(D-frac1Xright),(D-1),.$$
That is,
$$D,Biggl(fracexp(X)X,D,big(exp(-X),ybig)Biggr)=0,.$$
Consequently, for some constant $A$,
$$big(D,big(exp(-X),ybig)Big)(x)=-A,x,exp(-x),,$$
making
$$y(x)=exp(x),int,big(-A,x,exp(-x)big),textdx=A,(x+1)+B,exp(x),,$$
for some constant $B$.
answered Aug 2 at 8:25
Batominovski
22.9k22777
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Your work is correct up to the partial fractions
$$ fracv''v' = fracx^2+1x^2+x = 1 + frac1-xx(x+1) = 1 + frac1x - frac2x+1 $$
Integrating this gives
$$ ln(v') = x + ln x - 2ln(x+1) $$
$$ implies v' = fracxe^x(x+1)^2 = frace^xx+1-frace^x(x+1)^2 $$
This has the form $e^xf(x) + e^xf'(x) = (e^xf(x))'$, therefore
$$ v(x) = frace^xx+1 $$
I've ignored the integration constants, since they're already included in the general solution, which is
$$ u(x) = (x+1)(c_1 + c_2v(x)) = c_1(x+1) + c_2e^x $$
answered Jul 29 at 12:20
Dylan
11.4k31026
add a comment |Â
up vote
5
down vote
accepted
Your work is correct up to the partial fractions
$$ fracv''v' = fracx^2+1x^2+x = 1 + frac1-xx(x+1) = 1 + frac1x - frac2x+1 $$
Integrating this gives
$$ ln(v') = x + ln x - 2ln(x+1) $$
$$ implies v' = fracxe^x(x+1)^2 = frace^xx+1-frace^x(x+1)^2 $$
This has the form $e^xf(x) + e^xf'(x) = (e^xf(x))'$, therefore
$$ v(x) = frace^xx+1 $$
I've ignored the integration constants, since they're already included in the general solution, which is
$$ u(x) = (x+1)(c_1 + c_2v(x)) = c_1(x+1) + c_2e^x $$
answered Jul 29 at 12:20
Dylan
11.4k31026
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Your work is correct up to the partial fractions
$$ fracv''v' = fracx^2+1x^2+x = 1 + frac1-xx(x+1) = 1 + frac1x - frac2x+1 $$
Integrating this gives
$$ ln(v') = x + ln x - 2ln(x+1) $$
$$ implies v' = fracxe^x(x+1)^2 = frace^xx+1-frace^x(x+1)^2 $$
This has the form $e^xf(x) + e^xf'(x) = (e^xf(x))'$, therefore
$$ v(x) = frace^xx+1 $$
I've ignored the integration constants, since they're already included in the general solution, which is
$$ u(x) = (x+1)(c_1 + c_2v(x)) = c_1(x+1) + c_2e^x $$
answered Jul 29 at 12:20
Dylan
11.4k31026
Your work is correct up to the partial fractions
$$ fracv''v' = fracx^2+1x^2+x = 1 + frac1-xx(x+1) = 1 + frac1x - frac2x+1 $$
Integrating this gives
$$ ln(v') = x + ln x - 2ln(x+1) $$
$$ implies v' = fracxe^x(x+1)^2 = frace^xx+1-frace^x(x+1)^2 $$
This has the form $e^xf(x) + e^xf'(x) = (e^xf(x))'$, therefore
$$ v(x) = frace^xx+1 $$
I've ignored the integration constants, since they're already included in the general solution, which is
$$ u(x) = (x+1)(c_1 + c_2v(x)) = c_1(x+1) + c_2e^x $$
answered Jul 29 at 12:20
Dylan
11.4k31026
answered Jul 29 at 12:20
Dylan
11.4k31026
answered Jul 29 at 12:20
Dylan
11.4k31026
answered Jul 29 at 12:20
Dylan
11.4k31026
11.4k31026
add a comment |Â
add a comment |Â
up vote
5
down vote
$$xu''-(x+1)u'+u=0$$
$$x(u''-u')-(u'-u)=0$$
Let $v=u'-u$,
$$xv'-v=0$$
beginalign
xv' &= v \
fracdvdx &= fracvx \
ln v &= ln x + C \
v &= Ax \
u'-u &= Ax \
u &= dfrac1expleft(-int dxright)int Axexpleft(-int dxright) dx \
&= Ae^xint xe^-x dx \
&= Ae^x left[-xe^-x-int-e^-xdxright] \
&= Ae^x left[-(x+1)e^-x+Bright] \
&= k_1(x+1)+k_2e^x \
endalign
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
add a comment |Â
up vote
5
down vote
$$xu''-(x+1)u'+u=0$$
$$x(u''-u')-(u'-u)=0$$
Let $v=u'-u$,
$$xv'-v=0$$
beginalign
xv' &= v \
fracdvdx &= fracvx \
ln v &= ln x + C \
v &= Ax \
u'-u &= Ax \
u &= dfrac1expleft(-int dxright)int Axexpleft(-int dxright) dx \
&= Ae^xint xe^-x dx \
&= Ae^x left[-xe^-x-int-e^-xdxright] \
&= Ae^x left[-(x+1)e^-x+Bright] \
&= k_1(x+1)+k_2e^x \
endalign
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$xu''-(x+1)u'+u=0$$
$$x(u''-u')-(u'-u)=0$$
Let $v=u'-u$,
$$xv'-v=0$$
beginalign
xv' &= v \
fracdvdx &= fracvx \
ln v &= ln x + C \
v &= Ax \
u'-u &= Ax \
u &= dfrac1expleft(-int dxright)int Axexpleft(-int dxright) dx \
&= Ae^xint xe^-x dx \
&= Ae^x left[-xe^-x-int-e^-xdxright] \
&= Ae^x left[-(x+1)e^-x+Bright] \
&= k_1(x+1)+k_2e^x \
endalign
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
$$xu''-(x+1)u'+u=0$$
$$x(u''-u')-(u'-u)=0$$
Let $v=u'-u$,
$$xv'-v=0$$
beginalign
xv' &= v \
fracdvdx &= fracvx \
ln v &= ln x + C \
v &= Ax \
u'-u &= Ax \
u &= dfrac1expleft(-int dxright)int Axexpleft(-int dxright) dx \
&= Ae^xint xe^-x dx \
&= Ae^x left[-xe^-x-int-e^-xdxright] \
&= Ae^x left[-(x+1)e^-x+Bright] \
&= k_1(x+1)+k_2e^x \
endalign
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
edited Jul 29 at 12:19
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
answered Jul 29 at 12:05
Karn Watcharasupat
3,7992426
3,7992426
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
add a comment |Â
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
Very nice! I hadn't thought of this.
– Bell
Jul 29 at 12:13
add a comment |Â
up vote
1
down vote
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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-left(fracX+1Xright),D+frac1X=left(D-1-frac1X+frac1X+1right),left(D-frac1X+1right),.$$
First, let $v:=left(D-dfrac1X+1right),u$. Then, $left(D-1-dfrac1X+dfrac1X+1right),v=0$. Therefore,
$$v(x)=expBiggl(int,left(1+frac1x-frac1x+1right),textdxBiggr)=a,left(fracx,exp(x)x+1right)$$
for some constant $a$. Now, $v=left(D-dfrac1X+1right),u$ implies that $D,left(dfrac1X+1,uright)=dfrac1X+1,v$. Hence,
$$u(x)=(x+1),int,fracx,exp(x)(x+1)^2,textdx=(x+1),Biggl(aleft(fracexp(x)x+1right)+bBiggr)=a,exp(x)+b(x+1),,$$
for some constant $b$.
Interestingly, you can also write
$$D^2-left(fracX+1Xright),D+frac1X=left(D-frac1Xright),(D-1),.$$
That is,
$$D,Biggl(fracexp(X)X,D,big(exp(-X),ybig)Biggr)=0,.$$
Consequently, for some constant $A$,
$$big(D,big(exp(-X),ybig)Big)(x)=-A,x,exp(-x),,$$
making
$$y(x)=exp(x),int,big(-A,x,exp(-x)big),textdx=A,(x+1)+B,exp(x),,$$
for some constant $B$.
answered Aug 2 at 8:25
Batominovski
22.9k22777
add a comment |Â
up vote
1
down vote
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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-left(fracX+1Xright),D+frac1X=left(D-1-frac1X+frac1X+1right),left(D-frac1X+1right),.$$
First, let $v:=left(D-dfrac1X+1right),u$. Then, $left(D-1-dfrac1X+dfrac1X+1right),v=0$. Therefore,
$$v(x)=expBiggl(int,left(1+frac1x-frac1x+1right),textdxBiggr)=a,left(fracx,exp(x)x+1right)$$
for some constant $a$. Now, $v=left(D-dfrac1X+1right),u$ implies that $D,left(dfrac1X+1,uright)=dfrac1X+1,v$. Hence,
$$u(x)=(x+1),int,fracx,exp(x)(x+1)^2,textdx=(x+1),Biggl(aleft(fracexp(x)x+1right)+bBiggr)=a,exp(x)+b(x+1),,$$
for some constant $b$.
Interestingly, you can also write
$$D^2-left(fracX+1Xright),D+frac1X=left(D-frac1Xright),(D-1),.$$
That is,
$$D,Biggl(fracexp(X)X,D,big(exp(-X),ybig)Biggr)=0,.$$
Consequently, for some constant $A$,
$$big(D,big(exp(-X),ybig)Big)(x)=-A,x,exp(-x),,$$
making
$$y(x)=exp(x),int,big(-A,x,exp(-x)big),textdx=A,(x+1)+B,exp(x),,$$
for some constant $B$.
answered Aug 2 at 8:25
Batominovski
22.9k22777
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-left(fracX+1Xright),D+frac1X=left(D-1-frac1X+frac1X+1right),left(D-frac1X+1right),.$$
First, let $v:=left(D-dfrac1X+1right),u$. Then, $left(D-1-dfrac1X+dfrac1X+1right),v=0$. Therefore,
$$v(x)=expBiggl(int,left(1+frac1x-frac1x+1right),textdxBiggr)=a,left(fracx,exp(x)x+1right)$$
for some constant $a$. Now, $v=left(D-dfrac1X+1right),u$ implies that $D,left(dfrac1X+1,uright)=dfrac1X+1,v$. Hence,
$$u(x)=(x+1),int,fracx,exp(x)(x+1)^2,textdx=(x+1),Biggl(aleft(fracexp(x)x+1right)+bBiggr)=a,exp(x)+b(x+1),,$$
for some constant $b$.
Interestingly, you can also write
$$D^2-left(fracX+1Xright),D+frac1X=left(D-frac1Xright),(D-1),.$$
That is,
$$D,Biggl(fracexp(X)X,D,big(exp(-X),ybig)Biggr)=0,.$$
Consequently, for some constant $A$,
$$big(D,big(exp(-X),ybig)Big)(x)=-A,x,exp(-x),,$$
making
$$y(x)=exp(x),int,big(-A,x,exp(-x)big),textdx=A,(x+1)+B,exp(x),,$$
for some constant $B$.
answered Aug 2 at 8:25
Batominovski
22.9k22777
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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-left(fracX+1Xright),D+frac1X=left(D-1-frac1X+frac1X+1right),left(D-frac1X+1right),.$$
First, let $v:=left(D-dfrac1X+1right),u$. Then, $left(D-1-dfrac1X+dfrac1X+1right),v=0$. Therefore,
$$v(x)=expBiggl(int,left(1+frac1x-frac1x+1right),textdxBiggr)=a,left(fracx,exp(x)x+1right)$$
for some constant $a$. Now, $v=left(D-dfrac1X+1right),u$ implies that $D,left(dfrac1X+1,uright)=dfrac1X+1,v$. Hence,
$$u(x)=(x+1),int,fracx,exp(x)(x+1)^2,textdx=(x+1),Biggl(aleft(fracexp(x)x+1right)+bBiggr)=a,exp(x)+b(x+1),,$$
for some constant $b$.
Interestingly, you can also write
$$D^2-left(fracX+1Xright),D+frac1X=left(D-frac1Xright),(D-1),.$$
That is,
$$D,Biggl(fracexp(X)X,D,big(exp(-X),ybig)Biggr)=0,.$$
Consequently, for some constant $A$,
$$big(D,big(exp(-X),ybig)Big)(x)=-A,x,exp(-x),,$$
making
$$y(x)=exp(x),int,big(-A,x,exp(-x)big),textdx=A,(x+1)+B,exp(x),,$$
for some constant $B$.
answered Aug 2 at 8:25
Batominovski
22.9k22777
edited Aug 2 at 9:04
answered Aug 2 at 8:25
Batominovski
22.9k22777
answered Aug 2 at 8:25
Batominovski
22.9k22777
answered Aug 2 at 8:25
Batominovski
22.9k22777
22.9k22777
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4
There's nothing wrong with leaving the solution as an integral. Also, the last integral is the exponential integral
– Dylan
Jul 29 at 11:51
The answer provided is $u_2=Ae^x$. I'd like to know, if possible, how they got this answer
– Bell
Jul 29 at 11:54
1
If that's the case, then you must have done something wrong. I'd go back and check if your derived equation (second line) is correct
– Dylan
Jul 29 at 11:55
I will have another look, but I've read through it many times. I haven't found a mistake yet.
– Bell
Jul 29 at 11:58
4
Your partial fraction expansion has an error. The minus sign should be plus and the plus sign should be minus. Maybe that will fix the final answer.
– B. Goddard
Jul 29 at 12:08