Integration factor method with constants of integration The question at hand is a step in solving the manifolds of a non-linear dynamical system. However the part I need help with is a step using the integrating factor method.
We need to solve the 1st order ODE $$\frac{dy}{dt}-y=(x_0)^2e^{-2t}.$$
Me and my lecturer both agreed that the integration factor is $e^{-t}$, and working it out the way I usually do I got my answer: $$e^{-t}y(t)=\frac{{x_0}^2}{3}-\frac{{x_0}^2}{3}e^{-3t}.$$
However this is apparently wrong as my lecturer had an extra term, $y_0$, which I'm assuming is an integration constant but I cannot work out where it has come from. So his answer was $$e^{-t}y(t)=y_0+\int^{t}_{0}x_0^2e^{-3s}ds=y_0+\frac{{x_0}^2}{3}-\frac{{x_0}^2}{3}e^{-3t}.$$ It then goes on to have much importance in calculating the manifold so please can someone tell me how he obtained the $y_o$!
 A: Your differential equation is $$\frac{dy}{dt}-y=x_0^2e^{-2t}\tag1$$
Clearly, it is a first order linear differential equation and the integrating factor is $$\text{I.F.}~~=~e^{\int (-1)~dt}=e^{-t}$$
Now multiplying both side of $(1)$ by I.F. we have,
$$e^{-t}\frac{dy}{dt}-ye^{-t}=x_0^2e^{-3t}$$
$$\implies \frac{d}{dt}\left(ye^{-t}\right)=x_0^2e^{-3t}$$
$$\implies d\left(ye^{-t}\right)=x_0^2e^{-3t}dt\tag2$$
Integrating equation $(2)$ between the limit $~0~$ to $~t~$, we have
$$\left[ye^{-t}\right]_0^t=\int_0^t x_0^2e^{-3s}ds$$
$$\implies y(t)e^{-t}-y(0)e^{-0}=\int_0^t x_0^2e^{-3s}ds$$
$$\implies y(t)e^{-t}-y(0)=\int_0^t x_0^2e^{-3s}ds$$
$$\implies y(t)e^{-t}-y_0=\int_0^t x_0^2e^{-3s}ds$$
$$\implies y(t)e^{-t}=y_0+\int_0^t x_0^2e^{-3s}ds\tag3$$
$$\implies y(t)e^{-t}=y_0-\dfrac 13 x_0^2\left[e^{-3s}\right]_0^t$$
$$\implies y(t)e^{-t}=y_0-\dfrac 13 x_0^2\left[e^{-3t}-1\right]$$
$$\implies y(t)e^{-t}=y_0+\dfrac {x_0^2}{3}-\dfrac {x_0^2}{3} e^{-3t}\tag4$$
Combining equation $(3)$ and $(4)$ we have
$$y(t)e^{-t}=y_0+\int^{t}_{0}x_0^2e^{-3s}ds=y_0+\frac{{x_0}^2}{3}-\frac{{x_0}^2}{3}e^{-3t}.$$


For a first order linear differential equation $$\frac{dy}{dt}-p(t)y=q(t)~,$$ integrating factor is $$\text{I.F.}~~=~e^{\int p(t)~dt}$$
Also for more you can see Integrating Factor Method

