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- Thread starter naspek
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- #2

gabbagabbahey

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- #3

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still blur...

- #4

gabbagabbahey

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Call the FT of [itex]f(x)[/itex], [itex]F[f(x)]=\tilde{f}(k)[/itex]. Call the FT of [itex]u(x)[/itex], [itex]F[u(x)]=\tilde{u}(k)[/itex]. Your textbook likely tells you how to express the Fourier transform of [itex]-\frac{d^2u}{dx^2}[/itex] and [itex]a^2u(x)[/itex] in terms of [itex]\tilde{u}(k)[/itex]....do exactly that

- #5

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[tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)][/tex]

[tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+a^2F\left[u\right]=F[f(x)][/tex]

[tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-(-k^2\tilde{u}(k,y))\right]+a^2F\left[u\right]=F[f(x)][/tex]

- #6

gabbagabbahey

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This gives you an algebraic (instead of differential) equation you can solve for [itex]\tilde{u}[/itex]. To find [itex]u[/itex], you then take the inverse FT.

This is a very basic application of using FTs to solve DEs and I have a hard time believing that your textbook doesn't provide a similar example. You should really be in the habit of studying the relevant section of your text/notes before attempting homework problems that are based off of it.

- #7

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i got the general solution already.. howeverr.. how am i going to prove...

- #8

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[/itex], or [itex]k^2\tilde{u}(k)[/itex].

This gives you an algebraic (instead of differential) equation you can solve for [itex]\tilde{u}[/itex]. To find [itex]u[/itex], you then take the inverse FT.

This is a very basic application of using FTs to solve DEs and I have a hard time believing that your textbook doesn't provide a similar example. You should really be in the habit of studying the relevant section of your text/notes before attempting homework problems that are based off of it.

i've done the inverse..

F^-1[g^(k)] = g(x) = 1/ √2∏ ∫g^(k)e^ikx dk

.............................. = 1/ √2∏ ∫ 1/ (k^2 + a^2)e^ikx dk

...............................= √2∏ [(e^-a|x|) / 2a]

- #9

gabbagabbahey

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i've done the inverse..

F^-1[g^(k)] = g(x) = 1/ √2∏ ∫g^(k)e^ikx dk

.............................. = 1/ √2∏ ∫ 1/ (k^2 + a^2)e^ikx dk

...............................= √2∏ [(e^-a|x|) / 2a]

[tex]\tilde{u}(k)\neq\frac{1}{k^2+a^2}[/tex]

What happened to the FT of [itex]f(x)[/itex]?

- #10

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accidently double post

Last edited:

- #11

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ok.. this is my full solution...

am i got it right until here?

- #12

gabbagabbahey

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- #13

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inverse both side...

i get..

is that my general solution?

- #14

gabbagabbahey

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- #15

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by convolution theorem,

F^−1{F*G} = f ·g

correct?

so, i just substitute it..?

kinda confused

- #16

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ok.. i revised bout it already..

- #17

gabbagabbahey

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However, that's just part of the general solution. You can also add any function, [itex]u_h[/itex], which satisfies the corresponding homogeneous ODE, [itex]-\frac{d^2 u_h}{dx^2}+a^2u_h=0[/itex], to the [itex]u(x)[/itex] you've found, and the result will still satisfy your ODE. So, you need to find that homogeneous solution and add it to what you've found above in order to obtain the true general solution.

- #18

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However, that's just part of the general solution. You can also add any function, [itex]u_h[/itex], which satisfies the corresponding homogeneous ODE, [itex]-\frac{d^2 u_h}{dx^2}+a^2u_h=0[/itex], to the [itex]u(x)[/itex] you've found, and the result will still satisfy your ODE. So, you need to find that homogeneous solution and add it to what you've found above in order to obtain the true general solution.

should i solve the integration u(x) = 1/2a

- #19

gabbagabbahey

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- #20

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can u give me more clue.. i really can't get it..

which homogeneous equation i need to solve?

- #21

gabbagabbahey

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Re-read post #17

- #22

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[itex]

y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt

[/itex]

- #23

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after solving everything, i manage to get this equation..

\(\displaystyle u(x)= ∫- ∞ to 0

1/2a(x-y)^{5}e^{ay}dy+ ∫0 to ∞

(x-y)^{5}e^{-ay}dy

should i solve it to get the general solution?

because, i need to do integration by parts 5 times to solve it..\)

- #24

gabbagabbahey

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[itex]

y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt

[/itex]

[itex]u(x)=\frac{1}{2a}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt[/itex] is

Once you find the true general solution, you need only determine whether the integrals involved in exist/converge for your two [itex]f(x)[/itex]'s to know whether or not a solution exists.

- #25

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this is my general solution?

[itex]

u(x)=c_1 e^{-ax}+c_2 e^{ax}+y(x)

[/itex]

[itex]

u(x)=c_1 e^{-ax}+c_2 e^{ax}+y(x)

[/itex]

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