# Find the first two terms of the perturbation series solution to the IVP

Here is the equation for the initial value problem, $$y'=1+(1+\epsilon)y^2$$ with initial condition $$y(0)=1$$ with $$t>0$$ where $$0<\epsilon\ll1.$$. Find exact solution and compare approximation.

My attempt:

Suppose $$y=y_0+\epsilon y_1+O(\epsilon^2)$$ Then $$y_0'+\epsilon y_1'+O(\epsilon^2)=1+(1+\epsilon)(y_0+\epsilon y_1+O(\epsilon^2))^2$$, after distributing and combining like terms. $$\implies y_0'+\epsilon y_1'+O(\epsilon^2)=1+y_0^2+2\epsilon y_0y_1+\epsilon y_0^2+O(\epsilon^2)$$.

Grouping terms $$\\\implies (y_0'-y_0^2-1)+\epsilon(y_1'-2y_0y_1-y_0^2)+O(\epsilon^2)=0 \\$$.

Finally obtaining two sets of ODE's (This is where I stopped)$$\implies y_0'-y_0^2-1=0, \space y_1'-2y_0y_1-y_0^2=0 \implies \arctan(y_0)=t+C$$

Using initial value $$y(0)=1$$ we get $$C=\frac{\pi}{4}$$. So now we establish that $$y_0=\tan(t+\frac{\pi}{4})$$.

So this implies that, $$y_1'-2\tan(t+\frac{\pi}{4})y_1-\tan^2(t+\frac{\pi}{4})=0$$.

Am I going in the right direction? Solving the second set of ODE's just seems overly complicated so I used Wolfram Alpha.

Your last equation can be interpreted as $$y_1'+2\frac{u'}{u}y_1=\frac{u'^2}{u^2},~~~u(t)=\cos(t+\pi/4)$$ which has the surprisingly simple integrating factor $$u^2$$, so that $$\Bigl(\cos^2(t+\pi/4)y_1(t)\Bigr)'=\sin^2(t+\pi/4)=\frac12(1-\cos(2t+\pi/2))$$ This now should be easy to integrate.
What you did is correct. Then you have to solve $$y_1'-2\tan(t+\frac{\pi}{4})y_1-\tan^2(t+\frac{\pi}{4})=0$$. With initial condition you should obtain : $$\boxed{y'_1=-\frac12\tan(x+\frac{\pi}{4})+\frac14\left(1+\tan(x+\frac{\pi}{4})\right)(2x+1)}$$ It is easy to check in solving $$y'=1+(1+\epsilon)y^2$$ which is a separable ODE : $$t=\int\frac{dy}{1+(1+\epsilon)y^2}\quad\implies\quad t=\frac{1}{\sqrt{1+\epsilon}}\tan^{-1}(\sqrt{1+\epsilon}\:y)+c$$ And with condition $$y(0)=1$$ : $$y= \frac{1}{\sqrt{1+\epsilon}}\tan\left(\sqrt{1+\epsilon}\:t+\tan^{-1}(\sqrt{1+\epsilon}) \right)$$ Expending into power series of $$\epsilon$$ : $$y=\tan(t+\frac{\pi}{4})+\left(-\frac12\tan(x+\frac{\pi}{4})+\frac14\left(1+\tan(x+\frac{\pi}{4})\right)(2x+1)\right)\epsilon+...$$ This is consistent with the above results.