HELP, solve the differential equation, y''+ ty' + ty = 0Use Fourier transforms.

Question

solve the differential equation, y'' + ty' + ty = 0 Use Fourier transforms. 

Answer 1

There are a few ways to go about this. You can do this fairly quickly if you know the identities, or you can do it through integration using the definition. I will just speed through this using the identities for Fourier Transforms (FT).

To apply Fourier Transform (FT) to the equation y'' + ty' + ty = 0, you would look at it term by term as follows:

Recall: i² = -1

for the second order term y'', use the power property of FT:

F{ y''(t) } = ( iξ )² yˆ( ξ ) = -ξ² yˆ( ξ )

for the first order term, note that F { y' } =. i ξ yˆ (ξ)

you have a t in front, which requires you to use the identity

F { t f(t) } = i d/dξ ( fˆ (ξ) ) for some function f.

When you set up the integral $\int_{-\infty}^{\infty} t f(t) e^{-i\xi t} dt$, you notice that $t e^{-i\xi t}$ is related to the derivative of $e^{-i\xi t}$ with respect to $\xi$.

∫ t f(t) e^-iξt dt ( sorry the type setting for math on here is not good). you are integrating over the entire infinite domain R ( negative infinity to infinity).

Applying this to the first order term in your problem:

F{ t y'(t) } = i d\dξ ( F { y' } ) = i d\dξ ( i ξ yˆ (ξ) )

to differentiate, apply the product rule:

d\dξ ( i ξ yˆ (ξ) ) = i yˆ( ξ ) + i ξ d/dξ ( yˆ (ξ) )

multiplying by i yields: - yˆ( ξ ) - ξ d/dξ ( yˆ (ξ) )

Therefore, F{ t y'(t) } = - yˆ( ξ ) - ξ [ d/dξ ( yˆ (ξ) ) ]

for the last term, use the same identity F { t f(t) } = i d/dξ ( fˆ (ξ) )

Therefore F{ t y } = i d/dξ ( yˆ (ξ) )

putting the pieces together, you get

-ξ² yˆ( ξ ) - yˆ( ξ ) - ξ [ d/dξ ( yˆ (ξ) ) ] + i d/dξ ( yˆ (ξ) ) = 0

you can simply this as:

(i - ξ) d/dξ ( yˆ (ξ) ) = (ξ² + 1) yˆ (ξ)

for simplicity, rewrite as

(i - ξ) dyˆ/dξ = (ξ² + 1) yˆ

Notice that this is a separable ODE, rewrite it as

(i - ξ)/ yˆ dyˆ= (ξ² + 1) dξ

integrate both sides:

∫ 1/ yˆ dyˆ= ∫ (ξ² + 1)/(i - ξ) dξ

this gives you

ln | yˆ |

on the right side

for the left side, you can simplify what's inside the integral by factoring out

(ξ² + 1) = (ξ + i) ( ξ - i)

(i - ξ) = - (ξ - i)

this gives you (ξ + i) ( ξ - i) / - (ξ - i) = - (ξ + i)

integrating this gives you

-ξ/2 - i ξ + C

on the left hand side which you can simplify as

-(ξ/2 + i ξ) + C

so

ln | yˆ | = -(ξ/2 + i ξ) + C

to undo logs, use exponential:

ˆy = e^{-(ξ/2 + iξ) + C}

using exponent rules: e^{-(ξ/2 + iξ) + C =} e^{-(ξ/2 + iξ)} e^C

since e^C is just a constant, you can denote by some variable A = e^C

so ˆy = A e^{-(ξ/2 + iξ)}

Note that this is the solution in the frequency domain. you need to switch back to the original time t coordinates by doing inverse transform:

F^-1 { yˆ } = F^-1 { A e^{-(ξ/2 + iξ)} }

Note e^{-(ξ/2 + iξ)} = e^-ξ/2 e^-iξ

e^-ξ/2 is a Gaussian

For e^-iξ, recall the shift property of FT:

multiplying by e^^-iaξ in the frequency domain corresponds to a shift (t - a) in the time domain so

e^-iξ indicates a shift by 1, since a =1 here

therefore,

y = A e^{- [ (t-1)^2 ] / 2}