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Using Variation of Parameters to Find the General Solution of $u''-u=frac2e^x+1$
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I am trying to use variation of parameters to find the general solution to the inhomogeneous ODE
$$u''-u=frac2e^x+1$$
By computing the characteristic polynomial of the homogeneous equation $u''-u=0$, I have found $$u_H=c_1e^x+c_2e^-x c_1,c_2inmathbbR$$
Hence $u_1(x)=e^x, u_2(x)=e^-x$. So,
beginalign
v'_1(x)&=-fracu_2(x)f(x)W(x) \
v_1(x)&=int frace^-xe^x+1 dx \
&=-e^-x+ln|e^-x+1|+C_1
endalign
and similarly,
beginalign
v'_2(x)&=fracu_1(x)f(x)W(x) \
v_2(x)&=-int frace^xe^x+1 dx \
&=-ln|e^x+1|+C_2
endalign
Hence the general solution is given by
$$u(x)=v_1(x)u_1(x)+v_2(x)u_2(x)=left(-e^-x+ln|e^-x+1|+C_1right)e^x+left(-ln|e^x+1|+C_2right)e^-x$$
But this is incorret. Is my method valid?
integration differential-equations proof-verification
asked 20 hours ago
Bell
53612
 |Â
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up vote
0
down vote
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I am trying to use variation of parameters to find the general solution to the inhomogeneous ODE
$$u''-u=frac2e^x+1$$
By computing the characteristic polynomial of the homogeneous equation $u''-u=0$, I have found $$u_H=c_1e^x+c_2e^-x c_1,c_2inmathbbR$$
Hence $u_1(x)=e^x, u_2(x)=e^-x$. So,
beginalign
v'_1(x)&=-fracu_2(x)f(x)W(x) \
v_1(x)&=int frace^-xe^x+1 dx \
&=-e^-x+ln|e^-x+1|+C_1
endalign
and similarly,
beginalign
v'_2(x)&=fracu_1(x)f(x)W(x) \
v_2(x)&=-int frace^xe^x+1 dx \
&=-ln|e^x+1|+C_2
endalign
Hence the general solution is given by
$$u(x)=v_1(x)u_1(x)+v_2(x)u_2(x)=left(-e^-x+ln|e^-x+1|+C_1right)e^x+left(-ln|e^x+1|+C_2right)e^-x$$
But this is incorret. Is my method valid?
integration differential-equations proof-verification
asked 20 hours ago
Bell
53612
I have got$$u_p=-e^-xlog(e^x+1)+e^x(-e^-x+log(e^x+1))$$
– Dr. Sonnhard Graubner
20 hours ago
Indeed that is what I have determined. The solution that I have found has $$u_p=e^xln(1+e^-x)-xe^x-e^-xln(1+e^x)-1$$ which does not make much sense to me.
– Bell
20 hours ago
I think the $-xe^x$ is wrong
– Dr. Sonnhard Graubner
20 hours ago
I think so as well. I thought I might upload it to MSE as university literature (which is where the question is from) is very rarely found with mistakes. I might raise this with the lecturer. Thanks
– Bell
20 hours ago
You can tell me the result if you want.
– Dr. Sonnhard Graubner
20 hours ago
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to use variation of parameters to find the general solution to the inhomogeneous ODE
$$u''-u=frac2e^x+1$$
By computing the characteristic polynomial of the homogeneous equation $u''-u=0$, I have found $$u_H=c_1e^x+c_2e^-x c_1,c_2inmathbbR$$
Hence $u_1(x)=e^x, u_2(x)=e^-x$. So,
beginalign
v'_1(x)&=-fracu_2(x)f(x)W(x) \
v_1(x)&=int frace^-xe^x+1 dx \
&=-e^-x+ln|e^-x+1|+C_1
endalign
and similarly,
beginalign
v'_2(x)&=fracu_1(x)f(x)W(x) \
v_2(x)&=-int frace^xe^x+1 dx \
&=-ln|e^x+1|+C_2
endalign
Hence the general solution is given by
$$u(x)=v_1(x)u_1(x)+v_2(x)u_2(x)=left(-e^-x+ln|e^-x+1|+C_1right)e^x+left(-ln|e^x+1|+C_2right)e^-x$$
But this is incorret. Is my method valid?
integration differential-equations proof-verification
asked 20 hours ago
Bell
53612
I am trying to use variation of parameters to find the general solution to the inhomogeneous ODE
$$u''-u=frac2e^x+1$$
By computing the characteristic polynomial of the homogeneous equation $u''-u=0$, I have found $$u_H=c_1e^x+c_2e^-x c_1,c_2inmathbbR$$
Hence $u_1(x)=e^x, u_2(x)=e^-x$. So,
beginalign
v'_1(x)&=-fracu_2(x)f(x)W(x) \
v_1(x)&=int frace^-xe^x+1 dx \
&=-e^-x+ln|e^-x+1|+C_1
endalign
and similarly,
beginalign
v'_2(x)&=fracu_1(x)f(x)W(x) \
v_2(x)&=-int frace^xe^x+1 dx \
&=-ln|e^x+1|+C_2
endalign
Hence the general solution is given by
$$u(x)=v_1(x)u_1(x)+v_2(x)u_2(x)=left(-e^-x+ln|e^-x+1|+C_1right)e^x+left(-ln|e^x+1|+C_2right)e^-x$$
But this is incorret. Is my method valid?
integration differential-equations proof-verification
asked 20 hours ago
Bell
53612
edited 20 hours ago
asked 20 hours ago
Bell
53612
asked 20 hours ago
Bell
53612
asked 20 hours ago
Bell
53612
53612
I have got$$u_p=-e^-xlog(e^x+1)+e^x(-e^-x+log(e^x+1))$$
– Dr. Sonnhard Graubner
20 hours ago
Indeed that is what I have determined. The solution that I have found has $$u_p=e^xln(1+e^-x)-xe^x-e^-xln(1+e^x)-1$$ which does not make much sense to me.
– Bell
20 hours ago
I think the $-xe^x$ is wrong
– Dr. Sonnhard Graubner
20 hours ago
I think so as well. I thought I might upload it to MSE as university literature (which is where the question is from) is very rarely found with mistakes. I might raise this with the lecturer. Thanks
– Bell
20 hours ago
You can tell me the result if you want.
– Dr. Sonnhard Graubner
20 hours ago
 |Â
show 2 more comments
I have got$$u_p=-e^-xlog(e^x+1)+e^x(-e^-x+log(e^x+1))$$
– Dr. Sonnhard Graubner
20 hours ago
Indeed that is what I have determined. The solution that I have found has $$u_p=e^xln(1+e^-x)-xe^x-e^-xln(1+e^x)-1$$ which does not make much sense to me.
– Bell
20 hours ago
I think the $-xe^x$ is wrong
– Dr. Sonnhard Graubner
20 hours ago
I think so as well. I thought I might upload it to MSE as university literature (which is where the question is from) is very rarely found with mistakes. I might raise this with the lecturer. Thanks
– Bell
20 hours ago
You can tell me the result if you want.
– Dr. Sonnhard Graubner
20 hours ago
I have got$$u_p=-e^-xlog(e^x+1)+e^x(-e^-x+log(e^x+1))$$
– Dr. Sonnhard Graubner
20 hours ago
I have got$$u_p=-e^-xlog(e^x+1)+e^x(-e^-x+log(e^x+1))$$
– Dr. Sonnhard Graubner
20 hours ago
Indeed that is what I have determined. The solution that I have found has $$u_p=e^xln(1+e^-x)-xe^x-e^-xln(1+e^x)-1$$ which does not make much sense to me.
– Bell
20 hours ago
Indeed that is what I have determined. The solution that I have found has $$u_p=e^xln(1+e^-x)-xe^x-e^-xln(1+e^x)-1$$ which does not make much sense to me.
– Bell
20 hours ago
I think the $-xe^x$ is wrong
– Dr. Sonnhard Graubner
20 hours ago
I think the $-xe^x$ is wrong
– Dr. Sonnhard Graubner
20 hours ago
I think so as well. I thought I might upload it to MSE as university literature (which is where the question is from) is very rarely found with mistakes. I might raise this with the lecturer. Thanks
– Bell
20 hours ago
I think so as well. I thought I might upload it to MSE as university literature (which is where the question is from) is very rarely found with mistakes. I might raise this with the lecturer. Thanks
– Bell
20 hours ago
You can tell me the result if you want.
– Dr. Sonnhard Graubner
20 hours ago
You can tell me the result if you want.
– Dr. Sonnhard Graubner
20 hours ago
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
For particular solution, we can use differential operator method for $D(u''-u)=dfrac2e^x+1$ then
$$(D^2-1)u=frac2e^x+1$$
$$u=dfrac1D-1left(dfrac1D+1(frac2e^x+1)right)$$
let $v=dfrac1D+1(dfrac2e^x+1)$ then
$$e^xv'+e^xv=dfrac2e^xe^x+1$$
gives the function $v=e^-xintdfrac2e^xe^x+1dx=2e^-xln(e^x+1)$. Now
$$u=dfrac1D-1left(2e^-xln(e^x+1)right)$$
or
$$e^-xu'-e^-xu=2e^-2xln(e^x+1)$$
therefore
$$u=e^xint2e^-2xln(e^x+1)dx=colorbluee^xlnleft(e^-x+1right)-e^-xln left(e^x+1right)-1$$
answered 19 hours ago
user 108128
18.6k41544
add a comment |Â
up vote
2
down vote
First,
$$W(x)=detleft(beginbmatrixexp(x)&exp(-x)\exp(x)&-exp(-x)endbmatrixright)=-2,.$$
That is, with a proper constant of integration,
$$beginalignv_1(x)&=-int,fracu_2(x),f(x)W(x),textdx
\&=-exp(-x)+lnbig(1+exp(-x)big)=-exp(-x)-x+lnbig(1+exp(x)big),.endalign$$
Likewise, for an appropriate choice of the constant of integration, we have
$$v_2(x)=+int,fracu_1(x),f(x)W(x),textdx=-lnbig(1+exp(x)big),.$$
Thus, a particular solution is
$$beginalignu_p(x)&=v_1(x),u_1(x)+v_2(x),u_2(x)
\&=smallexp(+x),Big(-exp(-x)-x+lnbig(1+exp(x)big)Big)-exp(-x),lnbig(1+exp(x)big)
\
&=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big),.endalign$$
Hence, the general solutions are of the form
$$u(x)=u_p(x)+a,exp(+x)+b,exp(-x),,$$
where $a$ and $b$ are constants. My result agrees with you.
Alternatively, note that
$$fractextdtextdx,exp(x),big(u'(x)-u(x)big)=frac2,exp(x)exp(x)+1,.$$
That is,
$$u'(x)-u(x)=int,fracexp(x)exp(x)+1,textdx=2,exp(-x),lnbig(exp(x)+1big)-2A,exp(-x)$$
for some constant $A$. That is,
$$fractextdtextdx,exp(-x),u(x)=2,exp(-2x),lnbig(exp(x)+1big)-2A,exp(-2x),.$$
Ergo,
$$u(x)=exp(+x),int,2,exp(-2x),lnbig(exp(x)+1big),textdx+A,exp(-x),.$$
Using integration by parts, we obtain
$$int,2,exp(-2x),lnbig(exp(x)+1big),textdx=-exp(-x)-x+2exp(-x),sinh(x),lnbig(1+exp(x)big)+B$$
for some constant $B$, and so we conclude that
$$u(x)=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big)+A,exp(-x)+B,exp(+x),.$$
answered 19 hours ago
Batominovski
22.2k22675
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up vote
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$$u''-u=frac2e^x+1$$
$$u''-u'+u'-u=frac2e^x+1$$
Substitute $s=u'-u$
$$s'+s=frac2e^x+1$$
$$(se^x)'=frac2e^xe^x+1$$
Integrating
$$se^x=2ln(e^x+1)+K_1$$
$$(u'-u)e^-x=2e^-2xln(e^x+1)+K_1e^-2x$$
$$(ue^-x)'=2e^-2xln(e^x+1)+K_1e^-2x$$
After integration
$$u=2e^xint ln(e^x+1)e^-2xdx+K_1e^-x+K_2e^x$$
$$u=-ln(e^x+1)e^-x+e^xint frac e^-xe^x+1dx+K_1e^-x+K_2e^x$$
Finally
$$u(x)=-ln(e^x+1)e^-x+e^xln( e^-x+1)-1+K_1e^-x+K_2e^x$$
or
$$u(x)=ln(e^x+1)(-e^-x+e^x)-1-xe^x+K_1e^-x+K_2e^x$$
Edit
I think the $−xe^x$ is wrong
It's not wrong I think it comes from the
$$e^xln(e^-x+1)=e^xln (e^x+1)-xe^x$$
answered 19 hours ago
Isham
10.4k3729
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
1
ok @Bell have a nice day
– Isham
7 hours ago
 |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
2
down vote
For particular solution, we can use differential operator method for $D(u''-u)=dfrac2e^x+1$ then
$$(D^2-1)u=frac2e^x+1$$
$$u=dfrac1D-1left(dfrac1D+1(frac2e^x+1)right)$$
let $v=dfrac1D+1(dfrac2e^x+1)$ then
$$e^xv'+e^xv=dfrac2e^xe^x+1$$
gives the function $v=e^-xintdfrac2e^xe^x+1dx=2e^-xln(e^x+1)$. Now
$$u=dfrac1D-1left(2e^-xln(e^x+1)right)$$
or
$$e^-xu'-e^-xu=2e^-2xln(e^x+1)$$
therefore
$$u=e^xint2e^-2xln(e^x+1)dx=colorbluee^xlnleft(e^-x+1right)-e^-xln left(e^x+1right)-1$$
answered 19 hours ago
user 108128
18.6k41544
add a comment |Â
up vote
2
down vote
For particular solution, we can use differential operator method for $D(u''-u)=dfrac2e^x+1$ then
$$(D^2-1)u=frac2e^x+1$$
$$u=dfrac1D-1left(dfrac1D+1(frac2e^x+1)right)$$
let $v=dfrac1D+1(dfrac2e^x+1)$ then
$$e^xv'+e^xv=dfrac2e^xe^x+1$$
gives the function $v=e^-xintdfrac2e^xe^x+1dx=2e^-xln(e^x+1)$. Now
$$u=dfrac1D-1left(2e^-xln(e^x+1)right)$$
or
$$e^-xu'-e^-xu=2e^-2xln(e^x+1)$$
therefore
$$u=e^xint2e^-2xln(e^x+1)dx=colorbluee^xlnleft(e^-x+1right)-e^-xln left(e^x+1right)-1$$
answered 19 hours ago
user 108128
18.6k41544
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For particular solution, we can use differential operator method for $D(u''-u)=dfrac2e^x+1$ then
$$(D^2-1)u=frac2e^x+1$$
$$u=dfrac1D-1left(dfrac1D+1(frac2e^x+1)right)$$
let $v=dfrac1D+1(dfrac2e^x+1)$ then
$$e^xv'+e^xv=dfrac2e^xe^x+1$$
gives the function $v=e^-xintdfrac2e^xe^x+1dx=2e^-xln(e^x+1)$. Now
$$u=dfrac1D-1left(2e^-xln(e^x+1)right)$$
or
$$e^-xu'-e^-xu=2e^-2xln(e^x+1)$$
therefore
$$u=e^xint2e^-2xln(e^x+1)dx=colorbluee^xlnleft(e^-x+1right)-e^-xln left(e^x+1right)-1$$
answered 19 hours ago
user 108128
18.6k41544
For particular solution, we can use differential operator method for $D(u''-u)=dfrac2e^x+1$ then
$$(D^2-1)u=frac2e^x+1$$
$$u=dfrac1D-1left(dfrac1D+1(frac2e^x+1)right)$$
let $v=dfrac1D+1(dfrac2e^x+1)$ then
$$e^xv'+e^xv=dfrac2e^xe^x+1$$
gives the function $v=e^-xintdfrac2e^xe^x+1dx=2e^-xln(e^x+1)$. Now
$$u=dfrac1D-1left(2e^-xln(e^x+1)right)$$
or
$$e^-xu'-e^-xu=2e^-2xln(e^x+1)$$
therefore
$$u=e^xint2e^-2xln(e^x+1)dx=colorbluee^xlnleft(e^-x+1right)-e^-xln left(e^x+1right)-1$$
answered 19 hours ago
user 108128
18.6k41544
answered 19 hours ago
user 108128
18.6k41544
answered 19 hours ago
user 108128
18.6k41544
answered 19 hours ago
user 108128
18.6k41544
18.6k41544
add a comment |Â
add a comment |Â
up vote
2
down vote
First,
$$W(x)=detleft(beginbmatrixexp(x)&exp(-x)\exp(x)&-exp(-x)endbmatrixright)=-2,.$$
That is, with a proper constant of integration,
$$beginalignv_1(x)&=-int,fracu_2(x),f(x)W(x),textdx
\&=-exp(-x)+lnbig(1+exp(-x)big)=-exp(-x)-x+lnbig(1+exp(x)big),.endalign$$
Likewise, for an appropriate choice of the constant of integration, we have
$$v_2(x)=+int,fracu_1(x),f(x)W(x),textdx=-lnbig(1+exp(x)big),.$$
Thus, a particular solution is
$$beginalignu_p(x)&=v_1(x),u_1(x)+v_2(x),u_2(x)
\&=smallexp(+x),Big(-exp(-x)-x+lnbig(1+exp(x)big)Big)-exp(-x),lnbig(1+exp(x)big)
\
&=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big),.endalign$$
Hence, the general solutions are of the form
$$u(x)=u_p(x)+a,exp(+x)+b,exp(-x),,$$
where $a$ and $b$ are constants. My result agrees with you.
Alternatively, note that
$$fractextdtextdx,exp(x),big(u'(x)-u(x)big)=frac2,exp(x)exp(x)+1,.$$
That is,
$$u'(x)-u(x)=int,fracexp(x)exp(x)+1,textdx=2,exp(-x),lnbig(exp(x)+1big)-2A,exp(-x)$$
for some constant $A$. That is,
$$fractextdtextdx,exp(-x),u(x)=2,exp(-2x),lnbig(exp(x)+1big)-2A,exp(-2x),.$$
Ergo,
$$u(x)=exp(+x),int,2,exp(-2x),lnbig(exp(x)+1big),textdx+A,exp(-x),.$$
Using integration by parts, we obtain
$$int,2,exp(-2x),lnbig(exp(x)+1big),textdx=-exp(-x)-x+2exp(-x),sinh(x),lnbig(1+exp(x)big)+B$$
for some constant $B$, and so we conclude that
$$u(x)=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big)+A,exp(-x)+B,exp(+x),.$$
answered 19 hours ago
Batominovski
22.2k22675
add a comment |Â
up vote
2
down vote
First,
$$W(x)=detleft(beginbmatrixexp(x)&exp(-x)\exp(x)&-exp(-x)endbmatrixright)=-2,.$$
That is, with a proper constant of integration,
$$beginalignv_1(x)&=-int,fracu_2(x),f(x)W(x),textdx
\&=-exp(-x)+lnbig(1+exp(-x)big)=-exp(-x)-x+lnbig(1+exp(x)big),.endalign$$
Likewise, for an appropriate choice of the constant of integration, we have
$$v_2(x)=+int,fracu_1(x),f(x)W(x),textdx=-lnbig(1+exp(x)big),.$$
Thus, a particular solution is
$$beginalignu_p(x)&=v_1(x),u_1(x)+v_2(x),u_2(x)
\&=smallexp(+x),Big(-exp(-x)-x+lnbig(1+exp(x)big)Big)-exp(-x),lnbig(1+exp(x)big)
\
&=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big),.endalign$$
Hence, the general solutions are of the form
$$u(x)=u_p(x)+a,exp(+x)+b,exp(-x),,$$
where $a$ and $b$ are constants. My result agrees with you.
Alternatively, note that
$$fractextdtextdx,exp(x),big(u'(x)-u(x)big)=frac2,exp(x)exp(x)+1,.$$
That is,
$$u'(x)-u(x)=int,fracexp(x)exp(x)+1,textdx=2,exp(-x),lnbig(exp(x)+1big)-2A,exp(-x)$$
for some constant $A$. That is,
$$fractextdtextdx,exp(-x),u(x)=2,exp(-2x),lnbig(exp(x)+1big)-2A,exp(-2x),.$$
Ergo,
$$u(x)=exp(+x),int,2,exp(-2x),lnbig(exp(x)+1big),textdx+A,exp(-x),.$$
Using integration by parts, we obtain
$$int,2,exp(-2x),lnbig(exp(x)+1big),textdx=-exp(-x)-x+2exp(-x),sinh(x),lnbig(1+exp(x)big)+B$$
for some constant $B$, and so we conclude that
$$u(x)=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big)+A,exp(-x)+B,exp(+x),.$$
answered 19 hours ago
Batominovski
22.2k22675
add a comment |Â
up vote
2
down vote
up vote
2
down vote
First,
$$W(x)=detleft(beginbmatrixexp(x)&exp(-x)\exp(x)&-exp(-x)endbmatrixright)=-2,.$$
That is, with a proper constant of integration,
$$beginalignv_1(x)&=-int,fracu_2(x),f(x)W(x),textdx
\&=-exp(-x)+lnbig(1+exp(-x)big)=-exp(-x)-x+lnbig(1+exp(x)big),.endalign$$
Likewise, for an appropriate choice of the constant of integration, we have
$$v_2(x)=+int,fracu_1(x),f(x)W(x),textdx=-lnbig(1+exp(x)big),.$$
Thus, a particular solution is
$$beginalignu_p(x)&=v_1(x),u_1(x)+v_2(x),u_2(x)
\&=smallexp(+x),Big(-exp(-x)-x+lnbig(1+exp(x)big)Big)-exp(-x),lnbig(1+exp(x)big)
\
&=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big),.endalign$$
Hence, the general solutions are of the form
$$u(x)=u_p(x)+a,exp(+x)+b,exp(-x),,$$
where $a$ and $b$ are constants. My result agrees with you.
Alternatively, note that
$$fractextdtextdx,exp(x),big(u'(x)-u(x)big)=frac2,exp(x)exp(x)+1,.$$
That is,
$$u'(x)-u(x)=int,fracexp(x)exp(x)+1,textdx=2,exp(-x),lnbig(exp(x)+1big)-2A,exp(-x)$$
for some constant $A$. That is,
$$fractextdtextdx,exp(-x),u(x)=2,exp(-2x),lnbig(exp(x)+1big)-2A,exp(-2x),.$$
Ergo,
$$u(x)=exp(+x),int,2,exp(-2x),lnbig(exp(x)+1big),textdx+A,exp(-x),.$$
Using integration by parts, we obtain
$$int,2,exp(-2x),lnbig(exp(x)+1big),textdx=-exp(-x)-x+2exp(-x),sinh(x),lnbig(1+exp(x)big)+B$$
for some constant $B$, and so we conclude that
$$u(x)=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big)+A,exp(-x)+B,exp(+x),.$$
answered 19 hours ago
Batominovski
22.2k22675
First,
$$W(x)=detleft(beginbmatrixexp(x)&exp(-x)\exp(x)&-exp(-x)endbmatrixright)=-2,.$$
That is, with a proper constant of integration,
$$beginalignv_1(x)&=-int,fracu_2(x),f(x)W(x),textdx
\&=-exp(-x)+lnbig(1+exp(-x)big)=-exp(-x)-x+lnbig(1+exp(x)big),.endalign$$
Likewise, for an appropriate choice of the constant of integration, we have
$$v_2(x)=+int,fracu_1(x),f(x)W(x),textdx=-lnbig(1+exp(x)big),.$$
Thus, a particular solution is
$$beginalignu_p(x)&=v_1(x),u_1(x)+v_2(x),u_2(x)
\&=smallexp(+x),Big(-exp(-x)-x+lnbig(1+exp(x)big)Big)-exp(-x),lnbig(1+exp(x)big)
\
&=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big),.endalign$$
Hence, the general solutions are of the form
$$u(x)=u_p(x)+a,exp(+x)+b,exp(-x),,$$
where $a$ and $b$ are constants. My result agrees with you.
Alternatively, note that
$$fractextdtextdx,exp(x),big(u'(x)-u(x)big)=frac2,exp(x)exp(x)+1,.$$
That is,
$$u'(x)-u(x)=int,fracexp(x)exp(x)+1,textdx=2,exp(-x),lnbig(exp(x)+1big)-2A,exp(-x)$$
for some constant $A$. That is,
$$fractextdtextdx,exp(-x),u(x)=2,exp(-2x),lnbig(exp(x)+1big)-2A,exp(-2x),.$$
Ergo,
$$u(x)=exp(+x),int,2,exp(-2x),lnbig(exp(x)+1big),textdx+A,exp(-x),.$$
Using integration by parts, we obtain
$$int,2,exp(-2x),lnbig(exp(x)+1big),textdx=-exp(-x)-x+2exp(-x),sinh(x),lnbig(1+exp(x)big)+B$$
for some constant $B$, and so we conclude that
$$u(x)=-1-x,exp(x)+2,sinh(x),lnbig(1+exp(x)big)+A,exp(-x)+B,exp(+x),.$$
answered 19 hours ago
Batominovski
22.2k22675
edited 19 hours ago
answered 19 hours ago
Batominovski
22.2k22675
answered 19 hours ago
Batominovski
22.2k22675
answered 19 hours ago
Batominovski
22.2k22675
22.2k22675
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up vote
1
down vote
$$u''-u=frac2e^x+1$$
$$u''-u'+u'-u=frac2e^x+1$$
Substitute $s=u'-u$
$$s'+s=frac2e^x+1$$
$$(se^x)'=frac2e^xe^x+1$$
Integrating
$$se^x=2ln(e^x+1)+K_1$$
$$(u'-u)e^-x=2e^-2xln(e^x+1)+K_1e^-2x$$
$$(ue^-x)'=2e^-2xln(e^x+1)+K_1e^-2x$$
After integration
$$u=2e^xint ln(e^x+1)e^-2xdx+K_1e^-x+K_2e^x$$
$$u=-ln(e^x+1)e^-x+e^xint frac e^-xe^x+1dx+K_1e^-x+K_2e^x$$
Finally
$$u(x)=-ln(e^x+1)e^-x+e^xln( e^-x+1)-1+K_1e^-x+K_2e^x$$
or
$$u(x)=ln(e^x+1)(-e^-x+e^x)-1-xe^x+K_1e^-x+K_2e^x$$
Edit
I think the $−xe^x$ is wrong
It's not wrong I think it comes from the
$$e^xln(e^-x+1)=e^xln (e^x+1)-xe^x$$
answered 19 hours ago
Isham
10.4k3729
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
1
ok @Bell have a nice day
– Isham
7 hours ago
 |Â
show 1 more comment
up vote
1
down vote
$$u''-u=frac2e^x+1$$
$$u''-u'+u'-u=frac2e^x+1$$
Substitute $s=u'-u$
$$s'+s=frac2e^x+1$$
$$(se^x)'=frac2e^xe^x+1$$
Integrating
$$se^x=2ln(e^x+1)+K_1$$
$$(u'-u)e^-x=2e^-2xln(e^x+1)+K_1e^-2x$$
$$(ue^-x)'=2e^-2xln(e^x+1)+K_1e^-2x$$
After integration
$$u=2e^xint ln(e^x+1)e^-2xdx+K_1e^-x+K_2e^x$$
$$u=-ln(e^x+1)e^-x+e^xint frac e^-xe^x+1dx+K_1e^-x+K_2e^x$$
Finally
$$u(x)=-ln(e^x+1)e^-x+e^xln( e^-x+1)-1+K_1e^-x+K_2e^x$$
or
$$u(x)=ln(e^x+1)(-e^-x+e^x)-1-xe^x+K_1e^-x+K_2e^x$$
Edit
I think the $−xe^x$ is wrong
It's not wrong I think it comes from the
$$e^xln(e^-x+1)=e^xln (e^x+1)-xe^x$$
answered 19 hours ago
Isham
10.4k3729
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
1
ok @Bell have a nice day
– Isham
7 hours ago
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
$$u''-u=frac2e^x+1$$
$$u''-u'+u'-u=frac2e^x+1$$
Substitute $s=u'-u$
$$s'+s=frac2e^x+1$$
$$(se^x)'=frac2e^xe^x+1$$
Integrating
$$se^x=2ln(e^x+1)+K_1$$
$$(u'-u)e^-x=2e^-2xln(e^x+1)+K_1e^-2x$$
$$(ue^-x)'=2e^-2xln(e^x+1)+K_1e^-2x$$
After integration
$$u=2e^xint ln(e^x+1)e^-2xdx+K_1e^-x+K_2e^x$$
$$u=-ln(e^x+1)e^-x+e^xint frac e^-xe^x+1dx+K_1e^-x+K_2e^x$$
Finally
$$u(x)=-ln(e^x+1)e^-x+e^xln( e^-x+1)-1+K_1e^-x+K_2e^x$$
or
$$u(x)=ln(e^x+1)(-e^-x+e^x)-1-xe^x+K_1e^-x+K_2e^x$$
Edit
I think the $−xe^x$ is wrong
It's not wrong I think it comes from the
$$e^xln(e^-x+1)=e^xln (e^x+1)-xe^x$$
answered 19 hours ago
Isham
10.4k3729
$$u''-u=frac2e^x+1$$
$$u''-u'+u'-u=frac2e^x+1$$
Substitute $s=u'-u$
$$s'+s=frac2e^x+1$$
$$(se^x)'=frac2e^xe^x+1$$
Integrating
$$se^x=2ln(e^x+1)+K_1$$
$$(u'-u)e^-x=2e^-2xln(e^x+1)+K_1e^-2x$$
$$(ue^-x)'=2e^-2xln(e^x+1)+K_1e^-2x$$
After integration
$$u=2e^xint ln(e^x+1)e^-2xdx+K_1e^-x+K_2e^x$$
$$u=-ln(e^x+1)e^-x+e^xint frac e^-xe^x+1dx+K_1e^-x+K_2e^x$$
Finally
$$u(x)=-ln(e^x+1)e^-x+e^xln( e^-x+1)-1+K_1e^-x+K_2e^x$$
or
$$u(x)=ln(e^x+1)(-e^-x+e^x)-1-xe^x+K_1e^-x+K_2e^x$$
Edit
I think the $−xe^x$ is wrong
It's not wrong I think it comes from the
$$e^xln(e^-x+1)=e^xln (e^x+1)-xe^x$$
answered 19 hours ago
Isham
10.4k3729
edited 14 hours ago
answered 19 hours ago
Isham
10.4k3729
answered 19 hours ago
Isham
10.4k3729
answered 19 hours ago
Isham
10.4k3729
10.4k3729
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
1
ok @Bell have a nice day
– Isham
7 hours ago
 |Â
show 1 more comment
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
1
ok @Bell have a nice day
– Isham
7 hours ago
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
I don't quite understand. Could you please explain your edit.
– Bell
9 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
@Bell Is this more clear ? $$e^xln(e^-x+1)=e^xln(frac 1 e^x+1)=e^xln (e^x+1)-e^xln e^x$$
– Isham
8 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
Yep, I understand. Although the general solution in the book is $$u(x)=e^xleft(C_1+ln|1+e^-x|-xright)+e^-xleft(C_2-ln|1+e^x|right)-1$$ but this does not look equivalent to both of our answers. Does this answer look correct to you?
– Bell
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
there is a sign mistake for the exponential inside the ln fucntion No it dosent look correct to me
– Isham
7 hours ago
1
1
ok @Bell have a nice day
– Isham
7 hours ago
ok @Bell have a nice day
– Isham
7 hours ago
 |Â
show 1 more comment
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I have got$$u_p=-e^-xlog(e^x+1)+e^x(-e^-x+log(e^x+1))$$
– Dr. Sonnhard Graubner
20 hours ago
Indeed that is what I have determined. The solution that I have found has $$u_p=e^xln(1+e^-x)-xe^x-e^-xln(1+e^x)-1$$ which does not make much sense to me.
– Bell
20 hours ago
I think the $-xe^x$ is wrong
– Dr. Sonnhard Graubner
20 hours ago
I think so as well. I thought I might upload it to MSE as university literature (which is where the question is from) is very rarely found with mistakes. I might raise this with the lecturer. Thanks
– Bell
20 hours ago
You can tell me the result if you want.
– Dr. Sonnhard Graubner
20 hours ago