$$y''+(y'+2y)y'+2y^2=0 \tag{*}$$
1.when $ u=y^2 $ express $\frac{du}{dx}$,$\frac{d^2 u}{dx^2}$ in terms of $y,y',y''$
2.when satisfy initial condition $y(0)=1$, $y'(0)=1$ , find solution for differential equation in * above , but $0 \le x \le \frac{\pi}{2 \sqrt 3}$
attempt:
1. I wasn't sure about the substitution. it is said $u=y^2$
is the relation u -> y ->x?
$\frac{du}{dx}=2y \frac{dy}{dx}$
$\frac{d^2 u}{dx^2}=$$\frac{d}{dx}(2y \frac{dy}{dx})$
$\frac{d^2 u}{dx^2}=$$2(\frac{dy}{dx})^2+(2y \frac{d^2y}{dx})$
$\frac{d^2 u}{dx^2}=2(y')^2+2yy''$
is this right?
- number two I wasn't sure enough should I express $y''$ & $y'$ in u?
$\frac{u''-2(y')^2}{2y}=y''$ using what is known above, I tried but I got $\frac{u''}{2}-\frac{(u')^2}{2u}+u'+2u=0$ I'm sure I'm close to right here
substituting $$u(y)=y'(x)$$ then we get $$y^2(2u(y)^2+(yu'(y)+2)u(y)+2)=0$$ this can be written in the form $$\frac{du(y)}{dy}=-\frac{2\left(\frac{1}{u(y)}+u(y)+1\right)}{y}$$ can you solve this equation? the solution should be $$y(x)=c_2 e^{\frac{1}{2} \left(\log \left(\cos \left(\frac{1}{2} \sqrt{3} \left(2 x-c_1\right)\right)\right)-x\right)}$$