Question

Solve the differential equation with Fourier transformations: y'(t)+t*y(t)=0 ,y(0)=1

Answer 1

y+t•y=0

Find the integrating factor

Answer 2

Anytime, you are dealing (in general) with:
\(\large\color{#000000 }{ \displaystyle y'+q'(x)\cdot y=p(x)}\)

You are multipling the equation times \(\large\color{#000000 }{ e^{q(x)}}\)
all through the equation.

\(\large\color{#000000 }{ \displaystyle y'e^{q(x)}+q'(x)e^{q(x)}\cdot y=p(x)e^{q(x)}}\)

And the left side would become,
\(\large\color{#000000 }{ \displaystyle \frac{dy}{dx}\left[ye^{q(x)}\right]=p(x)e^{q(x)}}\)

Answer 3

And at that point you are capable of integrating both sides.

Answer 4

Is the integration factor e ^{x ^{2}/2}?

Answer 5

Yes,very good

Answer 6

So multiply
\(\large\color{#000000 }{ \displaystyle y'+yt=0}\)

times \(\large\color{#000000 }{ \displaystyle e^{t^2/2}}\)

Answer 7

\(\large\color{#000000 }{ \displaystyle y'e^{t^2/2}+ye^{t^2/2}t=0}\)

Answer 8

And then, work the product rule backwards (as I showed).

Answer 9

Okay I will try!

Answer 10

Sure, go ahead \(: )\)

Answer 11

(if you want to go ahead and integrate both sides, the integral of 0 is equal to C)

Answer 12

(integrate, of course, after you work the product rule backwards.)

Answer 13

Okay and then I can use Fourier transformations?

Answer 14

\[e ^{t ^{2}/2}*y=C\]

Answer 15

All you need is (will will consider that general case)
BUT, when p(x) is 0.

\(\large\color{#000000 }{ \displaystyle y'+w'(x)\cdot y=0}\)
And some initial point, lets call it \((4,2)\)

You are multipling the equation times \(\large\color{#000000 }{ e^{w(x)}}\)
all through the equation.

\(\large\color{#000000 }{ \displaystyle y'e^{w(x)}+w'(x)e^{w(x)}\cdot y=0}\)

And the left side would become,
\(\large\color{#000000 }{ \displaystyle \frac{dy}{dx}\left[ye^{w(x)}\right]=0}\)

Integrate both sides,
\(\large\color{#000000 }{ \displaystyle ye^{w(x)}=C}\)

Solve for y, \(\large\color{#000000 }{ \displaystyle y=\frac{C}{e^{w(x)}}}\)

Solve for C,

\(\large\color{#000000 }{ \displaystyle 2e^{w(4)}=C}\)

^^
this is constant.

So, the answer is (for this abstract problem)
\(\large\color{#000000 }{ \displaystyle y=\frac{2e^{w(4)}}{e^{w(x)}}}\)

Answer 16

I have \[y=\frac{ 1 }{ e ^{t ^{2}/2} }\]

Answer 17

Thank you!
I found it: \[F[y]=e ^{-t^2 /2} \]
So the answer is \[y= e^{-w^2/2}*sqrt(2*Pi)\]

Answer 18

????

Answer 19

What don't you understand @IrishBoy123 ?

Answer 20

where are the fourier transforms?

Answer 21

did i miss that bit?

Answer 22

Oh wait

Answer 23

as in

\[\large \hat{f}(\omega ) = \int_{-\infty}^\infty \,dx \qquad f(x)\ e^{- 2\pi i \omega \; x } \]

Answer 24

\[F[\frac{ 1 }{ \sqrt(2*\pi)} *e^{(-t^2)/2}]=e^{-w^{2}/2}\]
So \[F[e^{-t^{2}/2}]=\sqrt{2*\pi}*e^{-w^{2}/2}\]

Answer 25

@SolomonZelman can you verify my answer?