Mixing
Finding $g(x)$ given $g(x)= displaystyleint_0^1e^x+tg(t) dt +x $
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$$g(x)= displaystyleint_0^1e^x+tg(t) dt +x $$
How do I find $g(x)$ from the above functional equation?
Attempt(s):
- $dfracg(x)-xe^x= displaystyleint_0^1 e^t g(t)dt$
- Differentiated to get $g'(x)+x = g(x)+1$ and double differentiated to get $g''(x)+1 = g'(x)$
- Wrote g(-x) and added to get $g(x)+g(-x)$ to see if anything useful canbe extracted.
- Tried integration by parts.
Answer is:
$g(x)= left(dfrac23-e^2right)e^x + x$
calculus integration definite-integrals
asked Jul 28 at 17:53
Abcd
2,3151624
add a comment |Â
up vote
1
down vote
favorite
$$g(x)= displaystyleint_0^1e^x+tg(t) dt +x $$
How do I find $g(x)$ from the above functional equation?
Attempt(s):
- $dfracg(x)-xe^x= displaystyleint_0^1 e^t g(t)dt$
- Differentiated to get $g'(x)+x = g(x)+1$ and double differentiated to get $g''(x)+1 = g'(x)$
- Wrote g(-x) and added to get $g(x)+g(-x)$ to see if anything useful canbe extracted.
- Tried integration by parts.
Answer is:
$g(x)= left(dfrac23-e^2right)e^x + x$
calculus integration definite-integrals
asked Jul 28 at 17:53
Abcd
2,3151624
If you have gotten the differential equation, why not just solve that? It looks like a linear DE, should be rather easy to solve using Laplace Transforms.
– Pritt Balagopal
Jul 28 at 17:57
@PrittBalagopal We haven't been taught differential equations and Laplace transform.
– Abcd
Jul 28 at 17:57
Perhaps I am being dense based on these posted answers, but from $fracg(x) - xe^x = int_0^1e^tg(t)operatornamedt$. Doesn't this simply mean that the LHS must be constant for all $x$ and the form of $g(x)$ is clear from there?
– JessicaK
Jul 28 at 18:09
Yes @JessicaK Your comment was the most helpful to me . Thanks.
– Abcd
Jul 28 at 18:34
There is no need to differentiate. See my answer.
– JJacquelin
Aug 7 at 6:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$g(x)= displaystyleint_0^1e^x+tg(t) dt +x $$
How do I find $g(x)$ from the above functional equation?
Attempt(s):
- $dfracg(x)-xe^x= displaystyleint_0^1 e^t g(t)dt$
- Differentiated to get $g'(x)+x = g(x)+1$ and double differentiated to get $g''(x)+1 = g'(x)$
- Wrote g(-x) and added to get $g(x)+g(-x)$ to see if anything useful canbe extracted.
- Tried integration by parts.
Answer is:
$g(x)= left(dfrac23-e^2right)e^x + x$
calculus integration definite-integrals
asked Jul 28 at 17:53
Abcd
2,3151624
$$g(x)= displaystyleint_0^1e^x+tg(t) dt +x $$
How do I find $g(x)$ from the above functional equation?
Attempt(s):
- $dfracg(x)-xe^x= displaystyleint_0^1 e^t g(t)dt$
- Differentiated to get $g'(x)+x = g(x)+1$ and double differentiated to get $g''(x)+1 = g'(x)$
- Wrote g(-x) and added to get $g(x)+g(-x)$ to see if anything useful canbe extracted.
- Tried integration by parts.
Answer is:
$g(x)= left(dfrac23-e^2right)e^x + x$
calculus integration definite-integrals
asked Jul 28 at 17:53
Abcd
2,3151624
edited Jul 28 at 18:12
asked Jul 28 at 17:53
Abcd
2,3151624
asked Jul 28 at 17:53
Abcd
2,3151624
asked Jul 28 at 17:53
Abcd
2,3151624
2,3151624
If you have gotten the differential equation, why not just solve that? It looks like a linear DE, should be rather easy to solve using Laplace Transforms.
– Pritt Balagopal
Jul 28 at 17:57
@PrittBalagopal We haven't been taught differential equations and Laplace transform.
– Abcd
Jul 28 at 17:57
Perhaps I am being dense based on these posted answers, but from $fracg(x) - xe^x = int_0^1e^tg(t)operatornamedt$. Doesn't this simply mean that the LHS must be constant for all $x$ and the form of $g(x)$ is clear from there?
– JessicaK
Jul 28 at 18:09
Yes @JessicaK Your comment was the most helpful to me . Thanks.
– Abcd
Jul 28 at 18:34
There is no need to differentiate. See my answer.
– JJacquelin
Aug 7 at 6:32
add a comment |Â
If you have gotten the differential equation, why not just solve that? It looks like a linear DE, should be rather easy to solve using Laplace Transforms.
– Pritt Balagopal
Jul 28 at 17:57
@PrittBalagopal We haven't been taught differential equations and Laplace transform.
– Abcd
Jul 28 at 17:57
Perhaps I am being dense based on these posted answers, but from $fracg(x) - xe^x = int_0^1e^tg(t)operatornamedt$. Doesn't this simply mean that the LHS must be constant for all $x$ and the form of $g(x)$ is clear from there?
– JessicaK
Jul 28 at 18:09
Yes @JessicaK Your comment was the most helpful to me . Thanks.
– Abcd
Jul 28 at 18:34
There is no need to differentiate. See my answer.
– JJacquelin
Aug 7 at 6:32
If you have gotten the differential equation, why not just solve that? It looks like a linear DE, should be rather easy to solve using Laplace Transforms.
– Pritt Balagopal
Jul 28 at 17:57
If you have gotten the differential equation, why not just solve that? It looks like a linear DE, should be rather easy to solve using Laplace Transforms.
– Pritt Balagopal
Jul 28 at 17:57
@PrittBalagopal We haven't been taught differential equations and Laplace transform.
– Abcd
Jul 28 at 17:57
@PrittBalagopal We haven't been taught differential equations and Laplace transform.
– Abcd
Jul 28 at 17:57
Perhaps I am being dense based on these posted answers, but from $fracg(x) - xe^x = int_0^1e^tg(t)operatornamedt$. Doesn't this simply mean that the LHS must be constant for all $x$ and the form of $g(x)$ is clear from there?
– JessicaK
Jul 28 at 18:09
Perhaps I am being dense based on these posted answers, but from $fracg(x) - xe^x = int_0^1e^tg(t)operatornamedt$. Doesn't this simply mean that the LHS must be constant for all $x$ and the form of $g(x)$ is clear from there?
– JessicaK
Jul 28 at 18:09
Yes @JessicaK Your comment was the most helpful to me . Thanks.
– Abcd
Jul 28 at 18:34
Yes @JessicaK Your comment was the most helpful to me . Thanks.
– Abcd
Jul 28 at 18:34
There is no need to differentiate. See my answer.
– JJacquelin
Aug 7 at 6:32
There is no need to differentiate. See my answer.
– JJacquelin
Aug 7 at 6:32
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
3
down vote
accepted
hint
The solution of the differential equation
$$y'+x=y+1$$ satisfied by $g$ is
$$x+lambda e^x$$
with
$$lambda=int_0^1(x+lambda e^x)e^xdx$$
$$=1+fraclambda2(e^2-1)$$
thus
$$g(x)=x+frac2e^x(3-e^2)$$
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
add a comment |Â
up vote
3
down vote
Let $A:=displaystyle int_0^1,exp(t),g(t),textdt$. Then, the problem statement says that
$$g(x)=A,exp(x)+xtext for all xinmathbbR,.$$
Thus,
$$beginalign
A&=int_0^1,exp(t),g(t),textdt=int_0^1,exp(t),big(A,exp(t)+tbig),textdt
\
&=A,int_0^1,exp(2t),textdt+int_0^1,t,exp(t),textdt
\
&=fracA2,big(exp(2t)big)Big|^t=1_t=0+Biggl(big(t,exp(t)big)Big|_t=0^t=1-int_0^1,exp(t),textdtBiggr)
\
&=fracA2,left(texte^2-1right)+Biggl(texte-big(exp(t)big)Big|^t=1_t=0Biggr)=fracA2,left(texte^2-1right)+big(texte-(texte-1)big)
\&=fracA2,left(texte^2-1right)+1,.
endalign$$
This shows that
$$A=-frac2texte^2-3,,$$
whence
$$g(x)=-frac2,exp(x)texte^2-3+xtext for all xinmathbbR,,$$
which coincides with the answer key.
answered Jul 28 at 18:11
Batominovski
23k22777
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
add a comment |Â
up vote
1
down vote
You have came to the point where you got:
$$g'(x)-g(x)=1-x$$
This is a good news, you can solve differential equations of this form quite easily. Observing the LHS, you see that it's the (algebraic) sum of a derivative of the function, and the function itself, similar to the form obtained when you differentiate the multiplication of two functions. But how do you modify this equation to get such a result? The answer is integrating factor, which in this case is:
$$I.F = e^int-1dx$$
$$I.F = e^-x$$
Multiply the IF on both sides of the equation:
$$e^-xg'(x) - e^-xg(x) = e^-x(1-x)$$
You can see this conveniently becomes
$$fracddxe^-xg(x) = e^-x(1-x)$$
Hopefully you can take it from here.
answered Jul 28 at 18:05
Pritt Balagopal
517624
add a comment |Â
up vote
1
down vote
For the differential equation
$$g'(x)+x = g(x)+1$$
Rewrite it as
$$g'(x)-1 = g(x)-x$$
$$(g(x)-x)' = g(x)-x$$
$$int frac d(g(x)-x) g(x)-x=int dx=x+K$$
$$ln |g(x)-x|=x+K implies g(x)=Ke^x+x$$
answered Aug 7 at 1:05
Isham
10.5k3829
add a comment |Â
up vote
0
down vote
There is no need to differentiate.
$$fracg(x)-xe^x=displaystyleint_0^1e^tg(t) dt =C$$
Because the defined integral isn't function of $x$.
$$g(x)=x+Ce^x$$
$C=displaystyleint_0^1e^tleft(t+Ce^t right) dt = frace^2-12C+1quad$ gives $quad C=frac23-e^2$
$$g(x)=x+frac2e^x3-e^2$$
answered Aug 7 at 6:31
JJacquelin
39.8k21649
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
hint
The solution of the differential equation
$$y'+x=y+1$$ satisfied by $g$ is
$$x+lambda e^x$$
with
$$lambda=int_0^1(x+lambda e^x)e^xdx$$
$$=1+fraclambda2(e^2-1)$$
thus
$$g(x)=x+frac2e^x(3-e^2)$$
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
add a comment |Â
up vote
3
down vote
accepted
hint
The solution of the differential equation
$$y'+x=y+1$$ satisfied by $g$ is
$$x+lambda e^x$$
with
$$lambda=int_0^1(x+lambda e^x)e^xdx$$
$$=1+fraclambda2(e^2-1)$$
thus
$$g(x)=x+frac2e^x(3-e^2)$$
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
hint
The solution of the differential equation
$$y'+x=y+1$$ satisfied by $g$ is
$$x+lambda e^x$$
with
$$lambda=int_0^1(x+lambda e^x)e^xdx$$
$$=1+fraclambda2(e^2-1)$$
thus
$$g(x)=x+frac2e^x(3-e^2)$$
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
hint
The solution of the differential equation
$$y'+x=y+1$$ satisfied by $g$ is
$$x+lambda e^x$$
with
$$lambda=int_0^1(x+lambda e^x)e^xdx$$
$$=1+fraclambda2(e^2-1)$$
thus
$$g(x)=x+frac2e^x(3-e^2)$$
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
edited Jul 28 at 18:18
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
answered Jul 28 at 18:02
Salahamam_ Fatima
33.6k21229
33.6k21229
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
add a comment |Â
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
@Abcd $lambda=fracg(x)-xe^x=int_0^1e^xg(x)dx$ as you wrote.
– Salahamam_ Fatima
Jul 28 at 18:27
add a comment |Â
up vote
3
down vote
Let $A:=displaystyle int_0^1,exp(t),g(t),textdt$. Then, the problem statement says that
$$g(x)=A,exp(x)+xtext for all xinmathbbR,.$$
Thus,
$$beginalign
A&=int_0^1,exp(t),g(t),textdt=int_0^1,exp(t),big(A,exp(t)+tbig),textdt
\
&=A,int_0^1,exp(2t),textdt+int_0^1,t,exp(t),textdt
\
&=fracA2,big(exp(2t)big)Big|^t=1_t=0+Biggl(big(t,exp(t)big)Big|_t=0^t=1-int_0^1,exp(t),textdtBiggr)
\
&=fracA2,left(texte^2-1right)+Biggl(texte-big(exp(t)big)Big|^t=1_t=0Biggr)=fracA2,left(texte^2-1right)+big(texte-(texte-1)big)
\&=fracA2,left(texte^2-1right)+1,.
endalign$$
This shows that
$$A=-frac2texte^2-3,,$$
whence
$$g(x)=-frac2,exp(x)texte^2-3+xtext for all xinmathbbR,,$$
which coincides with the answer key.
answered Jul 28 at 18:11
Batominovski
23k22777
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
add a comment |Â
up vote
3
down vote
Let $A:=displaystyle int_0^1,exp(t),g(t),textdt$. Then, the problem statement says that
$$g(x)=A,exp(x)+xtext for all xinmathbbR,.$$
Thus,
$$beginalign
A&=int_0^1,exp(t),g(t),textdt=int_0^1,exp(t),big(A,exp(t)+tbig),textdt
\
&=A,int_0^1,exp(2t),textdt+int_0^1,t,exp(t),textdt
\
&=fracA2,big(exp(2t)big)Big|^t=1_t=0+Biggl(big(t,exp(t)big)Big|_t=0^t=1-int_0^1,exp(t),textdtBiggr)
\
&=fracA2,left(texte^2-1right)+Biggl(texte-big(exp(t)big)Big|^t=1_t=0Biggr)=fracA2,left(texte^2-1right)+big(texte-(texte-1)big)
\&=fracA2,left(texte^2-1right)+1,.
endalign$$
This shows that
$$A=-frac2texte^2-3,,$$
whence
$$g(x)=-frac2,exp(x)texte^2-3+xtext for all xinmathbbR,,$$
which coincides with the answer key.
answered Jul 28 at 18:11
Batominovski
23k22777
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $A:=displaystyle int_0^1,exp(t),g(t),textdt$. Then, the problem statement says that
$$g(x)=A,exp(x)+xtext for all xinmathbbR,.$$
Thus,
$$beginalign
A&=int_0^1,exp(t),g(t),textdt=int_0^1,exp(t),big(A,exp(t)+tbig),textdt
\
&=A,int_0^1,exp(2t),textdt+int_0^1,t,exp(t),textdt
\
&=fracA2,big(exp(2t)big)Big|^t=1_t=0+Biggl(big(t,exp(t)big)Big|_t=0^t=1-int_0^1,exp(t),textdtBiggr)
\
&=fracA2,left(texte^2-1right)+Biggl(texte-big(exp(t)big)Big|^t=1_t=0Biggr)=fracA2,left(texte^2-1right)+big(texte-(texte-1)big)
\&=fracA2,left(texte^2-1right)+1,.
endalign$$
This shows that
$$A=-frac2texte^2-3,,$$
whence
$$g(x)=-frac2,exp(x)texte^2-3+xtext for all xinmathbbR,,$$
which coincides with the answer key.
answered Jul 28 at 18:11
Batominovski
23k22777
Let $A:=displaystyle int_0^1,exp(t),g(t),textdt$. Then, the problem statement says that
$$g(x)=A,exp(x)+xtext for all xinmathbbR,.$$
Thus,
$$beginalign
A&=int_0^1,exp(t),g(t),textdt=int_0^1,exp(t),big(A,exp(t)+tbig),textdt
\
&=A,int_0^1,exp(2t),textdt+int_0^1,t,exp(t),textdt
\
&=fracA2,big(exp(2t)big)Big|^t=1_t=0+Biggl(big(t,exp(t)big)Big|_t=0^t=1-int_0^1,exp(t),textdtBiggr)
\
&=fracA2,left(texte^2-1right)+Biggl(texte-big(exp(t)big)Big|^t=1_t=0Biggr)=fracA2,left(texte^2-1right)+big(texte-(texte-1)big)
\&=fracA2,left(texte^2-1right)+1,.
endalign$$
This shows that
$$A=-frac2texte^2-3,,$$
whence
$$g(x)=-frac2,exp(x)texte^2-3+xtext for all xinmathbbR,,$$
which coincides with the answer key.
answered Jul 28 at 18:11
Batominovski
23k22777
answered Jul 28 at 18:11
Batominovski
23k22777
answered Jul 28 at 18:11
Batominovski
23k22777
answered Jul 28 at 18:11
Batominovski
23k22777
23k22777
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
add a comment |Â
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
As a remark, it is not true from the given condition that $$fracg(x)-xexp(x)=int_0^1,g(t),textdt,.$$ The correct version is $$fracg(x)-xexp(x)=int_0^1,colorredexp(t),g(t),textdt,.$$
– Batominovski
Jul 28 at 18:14
add a comment |Â
up vote
1
down vote
You have came to the point where you got:
$$g'(x)-g(x)=1-x$$
This is a good news, you can solve differential equations of this form quite easily. Observing the LHS, you see that it's the (algebraic) sum of a derivative of the function, and the function itself, similar to the form obtained when you differentiate the multiplication of two functions. But how do you modify this equation to get such a result? The answer is integrating factor, which in this case is:
$$I.F = e^int-1dx$$
$$I.F = e^-x$$
Multiply the IF on both sides of the equation:
$$e^-xg'(x) - e^-xg(x) = e^-x(1-x)$$
You can see this conveniently becomes
$$fracddxe^-xg(x) = e^-x(1-x)$$
Hopefully you can take it from here.
answered Jul 28 at 18:05
Pritt Balagopal
517624
add a comment |Â
up vote
1
down vote
You have came to the point where you got:
$$g'(x)-g(x)=1-x$$
This is a good news, you can solve differential equations of this form quite easily. Observing the LHS, you see that it's the (algebraic) sum of a derivative of the function, and the function itself, similar to the form obtained when you differentiate the multiplication of two functions. But how do you modify this equation to get such a result? The answer is integrating factor, which in this case is:
$$I.F = e^int-1dx$$
$$I.F = e^-x$$
Multiply the IF on both sides of the equation:
$$e^-xg'(x) - e^-xg(x) = e^-x(1-x)$$
You can see this conveniently becomes
$$fracddxe^-xg(x) = e^-x(1-x)$$
Hopefully you can take it from here.
answered Jul 28 at 18:05
Pritt Balagopal
517624
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have came to the point where you got:
$$g'(x)-g(x)=1-x$$
This is a good news, you can solve differential equations of this form quite easily. Observing the LHS, you see that it's the (algebraic) sum of a derivative of the function, and the function itself, similar to the form obtained when you differentiate the multiplication of two functions. But how do you modify this equation to get such a result? The answer is integrating factor, which in this case is:
$$I.F = e^int-1dx$$
$$I.F = e^-x$$
Multiply the IF on both sides of the equation:
$$e^-xg'(x) - e^-xg(x) = e^-x(1-x)$$
You can see this conveniently becomes
$$fracddxe^-xg(x) = e^-x(1-x)$$
Hopefully you can take it from here.
answered Jul 28 at 18:05
Pritt Balagopal
517624
You have came to the point where you got:
$$g'(x)-g(x)=1-x$$
This is a good news, you can solve differential equations of this form quite easily. Observing the LHS, you see that it's the (algebraic) sum of a derivative of the function, and the function itself, similar to the form obtained when you differentiate the multiplication of two functions. But how do you modify this equation to get such a result? The answer is integrating factor, which in this case is:
$$I.F = e^int-1dx$$
$$I.F = e^-x$$
Multiply the IF on both sides of the equation:
$$e^-xg'(x) - e^-xg(x) = e^-x(1-x)$$
You can see this conveniently becomes
$$fracddxe^-xg(x) = e^-x(1-x)$$
Hopefully you can take it from here.
answered Jul 28 at 18:05
Pritt Balagopal
517624
answered Jul 28 at 18:05
Pritt Balagopal
517624
answered Jul 28 at 18:05
Pritt Balagopal
517624
answered Jul 28 at 18:05
Pritt Balagopal
517624
517624
add a comment |Â
add a comment |Â
up vote
1
down vote
For the differential equation
$$g'(x)+x = g(x)+1$$
Rewrite it as
$$g'(x)-1 = g(x)-x$$
$$(g(x)-x)' = g(x)-x$$
$$int frac d(g(x)-x) g(x)-x=int dx=x+K$$
$$ln |g(x)-x|=x+K implies g(x)=Ke^x+x$$
answered Aug 7 at 1:05
Isham
10.5k3829
add a comment |Â
up vote
1
down vote
For the differential equation
$$g'(x)+x = g(x)+1$$
Rewrite it as
$$g'(x)-1 = g(x)-x$$
$$(g(x)-x)' = g(x)-x$$
$$int frac d(g(x)-x) g(x)-x=int dx=x+K$$
$$ln |g(x)-x|=x+K implies g(x)=Ke^x+x$$
answered Aug 7 at 1:05
Isham
10.5k3829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For the differential equation
$$g'(x)+x = g(x)+1$$
Rewrite it as
$$g'(x)-1 = g(x)-x$$
$$(g(x)-x)' = g(x)-x$$
$$int frac d(g(x)-x) g(x)-x=int dx=x+K$$
$$ln |g(x)-x|=x+K implies g(x)=Ke^x+x$$
answered Aug 7 at 1:05
Isham
10.5k3829
For the differential equation
$$g'(x)+x = g(x)+1$$
Rewrite it as
$$g'(x)-1 = g(x)-x$$
$$(g(x)-x)' = g(x)-x$$
$$int frac d(g(x)-x) g(x)-x=int dx=x+K$$
$$ln |g(x)-x|=x+K implies g(x)=Ke^x+x$$
answered Aug 7 at 1:05
Isham
10.5k3829
answered Aug 7 at 1:05
Isham
10.5k3829
answered Aug 7 at 1:05
Isham
10.5k3829
answered Aug 7 at 1:05
Isham
10.5k3829
10.5k3829
add a comment |Â
add a comment |Â
up vote
0
down vote
There is no need to differentiate.
$$fracg(x)-xe^x=displaystyleint_0^1e^tg(t) dt =C$$
Because the defined integral isn't function of $x$.
$$g(x)=x+Ce^x$$
$C=displaystyleint_0^1e^tleft(t+Ce^t right) dt = frace^2-12C+1quad$ gives $quad C=frac23-e^2$
$$g(x)=x+frac2e^x3-e^2$$
answered Aug 7 at 6:31
JJacquelin
39.8k21649
add a comment |Â
up vote
0
down vote
There is no need to differentiate.
$$fracg(x)-xe^x=displaystyleint_0^1e^tg(t) dt =C$$
Because the defined integral isn't function of $x$.
$$g(x)=x+Ce^x$$
$C=displaystyleint_0^1e^tleft(t+Ce^t right) dt = frace^2-12C+1quad$ gives $quad C=frac23-e^2$
$$g(x)=x+frac2e^x3-e^2$$
answered Aug 7 at 6:31
JJacquelin
39.8k21649
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is no need to differentiate.
$$fracg(x)-xe^x=displaystyleint_0^1e^tg(t) dt =C$$
Because the defined integral isn't function of $x$.
$$g(x)=x+Ce^x$$
$C=displaystyleint_0^1e^tleft(t+Ce^t right) dt = frace^2-12C+1quad$ gives $quad C=frac23-e^2$
$$g(x)=x+frac2e^x3-e^2$$
answered Aug 7 at 6:31
JJacquelin
39.8k21649
There is no need to differentiate.
$$fracg(x)-xe^x=displaystyleint_0^1e^tg(t) dt =C$$
Because the defined integral isn't function of $x$.
$$g(x)=x+Ce^x$$
$C=displaystyleint_0^1e^tleft(t+Ce^t right) dt = frace^2-12C+1quad$ gives $quad C=frac23-e^2$
$$g(x)=x+frac2e^x3-e^2$$
answered Aug 7 at 6:31
JJacquelin
39.8k21649
answered Aug 7 at 6:31
JJacquelin
39.8k21649
answered Aug 7 at 6:31
JJacquelin
39.8k21649
answered Aug 7 at 6:31
JJacquelin
39.8k21649
39.8k21649
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865445%2ffinding-gx-given-gx-displaystyle-int-01extgt-dt-x%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
If you have gotten the differential equation, why not just solve that? It looks like a linear DE, should be rather easy to solve using Laplace Transforms.
– Pritt Balagopal
Jul 28 at 17:57
@PrittBalagopal We haven't been taught differential equations and Laplace transform.
– Abcd
Jul 28 at 17:57
Perhaps I am being dense based on these posted answers, but from $fracg(x) - xe^x = int_0^1e^tg(t)operatornamedt$. Doesn't this simply mean that the LHS must be constant for all $x$ and the form of $g(x)$ is clear from there?
– JessicaK
Jul 28 at 18:09
Yes @JessicaK Your comment was the most helpful to me . Thanks.
– Abcd
Jul 28 at 18:34
There is no need to differentiate. See my answer.
– JJacquelin
Aug 7 at 6:32