I've seen similar questions here but none of them answered my problem. Or at least not the way that I would understand. I am trying to solve differential equation $$y'+t^2y=1.$$ I used the method of integrating factor - for this case $e^{t^3/3}$. By multiplying the equation i get $$e^{t^3/3}y'+e^{t^3/3}t^2y=e^{t^3/3}.$$ Now here is when I got confused. I should now integrate both sides (using indefinite integral?) which should result in $$c_1+e^{t^3/3}y=e^{t^3/3}.$$ So the solution should be $$y=e^{-t^3/3} \int e^{t^3/3} {dt} + c_2.$$ Now when I rewrite the integral in terms of a definite integral I get $$y=e^{-t^3/3} \int_{t_0}^{t} e^{s^3/3} {ds} + c_2.$$ My questions are first one - when do I change the variable in integral - is it because I change it from indefinite to definite? If so, why do I need to change the variable at all?

Second question - how do I manage the constants? Should I just keep it as $C$ or do I need to consider the initial condition too, since I change the integral from indefinite to definite? If so, how?



1 Answer 1


Your solution should be $$y(t)=e^{-t^3/3} \left(\int e^{t^3/3} {dt} + c_2\right).$$ (pay attention to the enclosing parentheses). In terms of a definite integral, it can be written as $$y(t)=e^{-t^3/3} \left(\int_{t_0}^t e^{s^3/3} {ds} + c_2\right)$$ where we renamed the variable inside the integral just to avoid confusion with the variable at the upper limit of the integration.

We may also find $c_2$ in terms of $y(t_0)$: by letting $t=t_0$ we get $$y(t_0)=e^{-t_0^3/3} \left(0 + c_2\right)\implies c_2=e^{t_0^3/3}y(t_0).$$

  • $\begingroup$ Thanks. You're right, I forgot the parentheses. Can you please write down more steps of how you got from the first equation to the other? Also what confusion would it make if i kept the variable $t$? Do you have any example? $\endgroup$
    – Lauren Sin
    Apr 16, 2020 at 11:49
  • 1
    $\begingroup$ See my edit. The confusion is that we use the same symbol $t$ for two different things: the upper limit and the variable of integration. If we are aware of such a distinction we may also write $$y(t)=e^{-t^3/3} \left(\int_{t_0}^t e^{t^3/3} {dt} + c_2\right).$$ The variable of integration is a dummy variable: mathworld.wolfram.com/DummyVariable.html $\endgroup$
    – Robert Z
    Apr 16, 2020 at 12:03

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