solve initial value problem (linear differential equation)
By Emma Terry •
To solve the initial value problem: $y' +2ty = e^{-t^2} , y(0) =y_0$
In which initial value $y_0$ comes from condition $y(2)=0$
Well, at least, I managed to solve the equation and find the general solution which is $y=te^{-t^2} + Ce^{-t^2}$. I think? but idk how to solve it with particular with initial value :)
ty for help etc :)
$\endgroup$ 31 Answer
$\begingroup$We have:
$$y(t) = e^{-t^2} ( c_1 + t )$$
At $t_0 = 2, y(2) = 0$, so we have:
$$y(2) = e^{-4}(c_1 + 2) = 0 \implies c_1 = -2$$
Hence
$$y(t) = e^{-t^2/2} (t - 2)$$
$\endgroup$ 0
