anonymous
  • anonymous
Solve the differential equation with Fourier transformations: y'(t)+t*y(t)=0 ,y(0)=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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SolomonZelman
  • SolomonZelman
y+t•y=0 Find the integrating factor
SolomonZelman
  • SolomonZelman
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, \(\large\color{#000000 }{ \displaystyle \frac{dy}{dx}\left[ye^{q(x)}\right]=p(x)e^{q(x)}}\)
SolomonZelman
  • SolomonZelman
And at that point you are capable of integrating both sides.

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anonymous
  • anonymous
Is the integration factor e ^{x ^{2}/2}?
SolomonZelman
  • SolomonZelman
Yes,very good
SolomonZelman
  • SolomonZelman
So multiply \(\large\color{#000000 }{ \displaystyle y'+yt=0}\) times \(\large\color{#000000 }{ \displaystyle e^{t^2/2}}\)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle y'e^{t^2/2}+ye^{t^2/2}t=0}\)
SolomonZelman
  • SolomonZelman
And then, work the product rule backwards (as I showed).
anonymous
  • anonymous
Okay I will try!
SolomonZelman
  • SolomonZelman
Sure, go ahead \(: )\)
SolomonZelman
  • SolomonZelman
(if you want to go ahead and integrate both sides, the integral of 0 is equal to C)
SolomonZelman
  • SolomonZelman
(integrate, of course, after you work the product rule backwards.)
anonymous
  • anonymous
Okay and then I can use Fourier transformations?
anonymous
  • anonymous
\[e ^{t ^{2}/2}*y=C\]
SolomonZelman
  • SolomonZelman
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) \(\large\color{#000000 }{ \displaystyle y=\frac{2e^{w(4)}}{e^{w(x)}}}\)
anonymous
  • anonymous
I have \[y=\frac{ 1 }{ e ^{t ^{2}/2} }\]
anonymous
  • anonymous
Thank you! I found it: \[F[y]=e ^{-t^2 /2} \] So the answer is \[y= e^{-w^2/2}*sqrt(2*Pi)\]
IrishBoy123
  • IrishBoy123
????
anonymous
  • anonymous
What don't you understand @IrishBoy123 ?
IrishBoy123
  • IrishBoy123
where are the fourier transforms?
IrishBoy123
  • IrishBoy123
did i miss that bit?
anonymous
  • anonymous
Oh wait
IrishBoy123
  • IrishBoy123
as in \[\large \hat{f}(\omega ) = \int_{-\infty}^\infty \,dx \qquad f(x)\ e^{- 2\pi i \omega \; x } \]
anonymous
  • anonymous
\[F[\frac{ 1 }{ \sqrt(2*\pi)} *e^{(-t^2)/2}]=e^{-w^{2}/2}\] So \[F[e^{-t^{2}/2}]=\sqrt{2*\pi}*e^{-w^{2}/2}\]
anonymous
  • anonymous
@SolomonZelman can you verify my answer?

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