Second order ODE solution The task is to find a solution of second order ODE:
$$
2xx''-3(x')^2=4x^2; x(0) = 1; x'(0) = 0
$$
Now, I tried:
$$
x'' = 2x + \frac{3(x')^2}{2x}
$$
here we substitute $u(x) = x'$ to obtain:
$$
uu'=2x+\frac{3}{2}\frac{u^2}{x} \\
u' = 2\frac{x}{u} + \frac{3u}{2x}
$$
here I substitute $t = \frac{x}{u}$ to get
$$
t'=\frac{2t+\frac{3}{2}\frac{1}{t} - t}{u} \\
\frac{dt}{du} = \frac{t+\frac{3}{2t}}{u} \\
\frac{du}{u} = \frac{dt}{t + \frac{3}{2t}}
$$
integrate both sides
$$
\ln u = \frac{1}{2}\ln{(2t^2 + 3)} + c \\
u = \sqrt{2t^2+3}e^c \\
x' = e^c\sqrt{2(\frac{x}{x'})^2 + 3}
$$
Here I got stuck and I think I made a mistake somewhere, because I can't use $x'(0)=1$ and $x(0)=0$ to get the constant $c$. Could you help me?
 A: Let me work with $y(x)$, then the ODE is
$$yy''-(4y^2+3y'^2)=0~~~~(1)$$
Let $y'=p, y''=\frac{dp}{dx}=\frac{dp}{dy} \frac{dy}{dx}=p\frac{dp}{dy}$
Then (1) becomes
$$yp\frac{dp}{dy}-(4y^2+3p^2)=0 \implies ypdp-(4y^2+3p^2)dy=0~~~~(2)$$
The integrating factor for this in-exact ODE is
$$\mu=\exp[\int \frac{-6p-p}{py}dy=y^{-7}$$
The ODE (2) becomes
$$py^{-6} dp-y^{-7}(4y^2+3p^2) dy=0$$
Its solution i1
$$\int y^{-6}  p dp-\int 4y^{-5}dy=C \implies \frac{p^2}{2y^6}+\frac{1}{y^4}-C=0$$
$$\implies p =\frac{dy}{dx}=\pm y\sqrt{2} \sqrt{Cy^4-1} \implies C=1$$ $$ \implies \int \frac{dy}{y\sqrt{y^4-1}}=\pm \int \sqrt{2} dx+B \implies -\frac{1}{2} \tan^{-1}\frac{1}{\sqrt{y^4-1}}= \pm \sqrt{2} x+B$$
$$\implies \tan^{-1}\frac{1}{\sqrt{y^4-1}}=\mp 2x\sqrt{2}+D \implies D=\pi/2$$
$$ \implies y(x)=\sqrt{\sec 2x\sqrt{2}}$$
A: $$2xx''-3(x')^2=4x^2; x(0) = 1; x'(0) = 0$$
Substitute $p=x'$ and $p'=\dfrac {dp}{dx}$:
$$2xp'p-3p^2=4x^2$$
$$p'-\dfrac 3 {2x}p=\dfrac {2x}p$$
This is simply Bernoulli's differential equation.
Your substitution should be: 
$$u=p^2 \implies \dfrac {du}{dx}=2pp'$$
This substitution change the DE into a linear first order DE.
$$\dfrac {u'}2-\dfrac 3 {2x}
u= {2x}$$
