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 :)

  • $\begingroup$ integrating factor to do what? if you mean solve the differential equation I already did that by multiplying both sides with e^(t^2) and then I got (y*e(^t^2))' dy =1dt and then I multiply with e^-t^2 to get y by itself so I get y=te^(-t^2) + Ce^(-t^2) :) its the substition part when im supposed to find c or whatever, that is confusing with this y_0 stuff :P $\endgroup$ – wellalright Apr 1 '14 at 1:45
  • $\begingroup$ Is it $y_0 = y(0)$ or $y_0 = y(2)$? $\endgroup$ – Graham Kemp Apr 1 '14 at 1:59
  • $\begingroup$ which initial valye y_0 comes from y(2)=0 that should be right.. sorry lol :/ $\endgroup$ – wellalright Apr 1 '14 at 2:00

We have:

$$y(t) = e^{-t^2} ( c_1 + t )$$

At $t_0 = 2, y(2) = 0$, so we have:

$$y(2) = e^{-4}(c_1 + 2) = 0 \implies c_1 = -2$$


$$y(t) = e^{-t^2/2} (t - 2)$$


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