Veryfing ODE for complicated y(t)

[rdioface]

Homework Statement

For the differential equation, verify (by differentiation and substitution) that the given function y(t) is a solution.

Homework Equations

$y' - 4ty = 1$

$y(t) = \int_{0}^{t} e^{-2(s^{2}-t^{2})} ds$

The Attempt at a Solution

I attempted to take $\frac{d}{dt}$ of y(t) as usual but
1. if I do not try bringing the $\frac{d}{dt}$ inside the integral I can do nothing because there is no elementary antiderivative of y(t).
2. if I do bring the $\frac{d}{dt}$ inside the integral, I can use the chain rule to get
$y(t) = (-2) \int_{0}^{t} (-2t) e^{-2(s^{2}-t^{2})} ds$
but since my variable of integration is ds not dt, this doesn't allow me to use a u-substitution as I had hoped nor can I think of a way to relate ds and dt.

More or less I do not know how to take $\frac{d}{dt}$ of y(t) and I do not know any other ways to solve the problem.

 

[τheory]

Do you have to do it that way? Because it looks like you can solve it linearly.

 

[rdioface]

Yes, the problem asks to verify that y(t) is a solution to the differential equation y' + 4ty = 1.

 

[Dick]

$(-2) \int_{0}^{t} (-2t) e^{-2(s^{2}-t^{2})} ds$
but since my variable of integration is ds not dt, this doesn't allow me to use a u-substitution as I had hoped nor can I think of a way to relate ds and dt.

I don't think you have to be able to do the integral to identify that expression as $4 t y(t)$. Since you are integrating ds you can factor out the t.

 

[rdioface]

I don't think you have to be able to do the integral to identify that expression as $4 t y(t)$. Since you are integrating ds you can factor out the t.

facepalm.jpg

Thanks!