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

Question

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.

Answer 1

I see no error in your derivation.

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.

Answer 2

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.