Solving ODE with Fourier Transform: $u(x) - u''(x) = x^2$
Solve:
$$u(x) - u''(x) = x^2$$
You can use:
$$
mathcal{F} { e^{-a|x|} } = frac{2a}{a^2 + s^2}
$$
I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
differential-equations fourier-transform
asked Dec 5 '18 at 12:51
BethBeth
111
add a comment |
Solve:
$$u(x) - u''(x) = x^2$$
You can use:
$$
mathcal{F} { e^{-a|x|} } = frac{2a}{a^2 + s^2}
$$
I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
differential-equations fourier-transform
asked Dec 5 '18 at 12:51
BethBeth
111
Welcome to Math.SE! What have you tried for the solution?
– rafa11111
Dec 5 '18 at 12:57
1
@rafa11111 I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
– Beth
Dec 5 '18 at 13:01
1
@Beth: insert that comment in the question.
– Yves Daoust
Dec 5 '18 at 13:04
You need to take the Fourier transform of both members.
– Yves Daoust
Dec 5 '18 at 13:05
add a comment |
Solve:
$$u(x) - u''(x) = x^2$$
You can use:
$$
mathcal{F} { e^{-a|x|} } = frac{2a}{a^2 + s^2}
$$
I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
differential-equations fourier-transform
asked Dec 5 '18 at 12:51
BethBeth
111
Solve:
$$u(x) - u''(x) = x^2$$
You can use:
$$
mathcal{F} { e^{-a|x|} } = frac{2a}{a^2 + s^2}
$$
I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
differential-equations fourier-transform
differential-equations fourier-transform
asked Dec 5 '18 at 12:51
BethBeth
111
asked Dec 5 '18 at 12:51
BethBeth
111
edited Dec 5 '18 at 13:06
Beth
asked Dec 5 '18 at 12:51
BethBeth
111
asked Dec 5 '18 at 12:51
BethBeth
111
asked Dec 5 '18 at 12:51
BethBeth
111
111
Welcome to Math.SE! What have you tried for the solution?
– rafa11111
Dec 5 '18 at 12:57
1
@rafa11111 I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
– Beth
Dec 5 '18 at 13:01
1
@Beth: insert that comment in the question.
– Yves Daoust
Dec 5 '18 at 13:04
You need to take the Fourier transform of both members.
– Yves Daoust
Dec 5 '18 at 13:05
add a comment |
Welcome to Math.SE! What have you tried for the solution?
– rafa11111
Dec 5 '18 at 12:57
1
@rafa11111 I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
– Beth
Dec 5 '18 at 13:01
1
@Beth: insert that comment in the question.
– Yves Daoust
Dec 5 '18 at 13:04
You need to take the Fourier transform of both members.
– Yves Daoust
Dec 5 '18 at 13:05
Welcome to Math.SE! What have you tried for the solution?
– rafa11111
Dec 5 '18 at 12:57
Welcome to Math.SE! What have you tried for the solution?
– rafa11111
Dec 5 '18 at 12:57
1
1
@rafa11111 I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
– Beth
Dec 5 '18 at 13:01
@rafa11111 I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
– Beth
Dec 5 '18 at 13:01
1
1
@Beth: insert that comment in the question.
– Yves Daoust
Dec 5 '18 at 13:04
@Beth: insert that comment in the question.
– Yves Daoust
Dec 5 '18 at 13:04
You need to take the Fourier transform of both members.
– Yves Daoust
Dec 5 '18 at 13:05
You need to take the Fourier transform of both members.
– Yves Daoust
Dec 5 '18 at 13:05
add a comment |
1 Answer
1
active
oldest
votes
You probably are to use that
$$
x^2=left.frac{partial^2}{partial a^2}e^{-a|x|}right|_{a=0}
$$
and consider first the solution of
$$
u_a(x)-u_a''(x)=e^{-a|x|}
$$
to then compute
$$
u=left.frac{partial^2}{partial a^2}u_a(x)right|_{a=0}.
$$
Check against the solution you get via undetermined coefficients, $u_p=ax^2+bx+c$.
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You probably are to use that
$$
x^2=left.frac{partial^2}{partial a^2}e^{-a|x|}right|_{a=0}
$$
and consider first the solution of
$$
u_a(x)-u_a''(x)=e^{-a|x|}
$$
to then compute
$$
u=left.frac{partial^2}{partial a^2}u_a(x)right|_{a=0}.
$$
Check against the solution you get via undetermined coefficients, $u_p=ax^2+bx+c$.
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
add a comment |
You probably are to use that
$$
x^2=left.frac{partial^2}{partial a^2}e^{-a|x|}right|_{a=0}
$$
and consider first the solution of
$$
u_a(x)-u_a''(x)=e^{-a|x|}
$$
to then compute
$$
u=left.frac{partial^2}{partial a^2}u_a(x)right|_{a=0}.
$$
Check against the solution you get via undetermined coefficients, $u_p=ax^2+bx+c$.
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
add a comment |
You probably are to use that
$$
x^2=left.frac{partial^2}{partial a^2}e^{-a|x|}right|_{a=0}
$$
and consider first the solution of
$$
u_a(x)-u_a''(x)=e^{-a|x|}
$$
to then compute
$$
u=left.frac{partial^2}{partial a^2}u_a(x)right|_{a=0}.
$$
Check against the solution you get via undetermined coefficients, $u_p=ax^2+bx+c$.
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
You probably are to use that
$$
x^2=left.frac{partial^2}{partial a^2}e^{-a|x|}right|_{a=0}
$$
and consider first the solution of
$$
u_a(x)-u_a''(x)=e^{-a|x|}
$$
to then compute
$$
u=left.frac{partial^2}{partial a^2}u_a(x)right|_{a=0}.
$$
Check against the solution you get via undetermined coefficients, $u_p=ax^2+bx+c$.
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
answered Dec 5 '18 at 13:16
LutzLLutzL
56.5k42054
56.5k42054
add a comment |
add a comment |
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Welcome to Math.SE! What have you tried for the solution?
– rafa11111
Dec 5 '18 at 12:57
1
@rafa11111 I am new to Fourier transforms. I understand that limits have to be used but don't know how to start. Any help will be appreciated.
– Beth
Dec 5 '18 at 13:01
1
@Beth: insert that comment in the question.
– Yves Daoust
Dec 5 '18 at 13:04
You need to take the Fourier transform of both members.
– Yves Daoust
Dec 5 '18 at 13:05