Помогите найти решение задачи Коши

$y''+18sin(y)cos^3(y)=0$

замена: $y'=p(y)$
$y''=pp'$

$pp'+18sin(y)cos^3(y)=0$

$pdp=-18sin(y)cos^3(y)dy$

$\frac{p^2}{2}=-18\int\limits {sin(y)cos^3(y)} \, dx$

$\frac{p^2}{2}=18\int\limits {cos^3(y)} \, d(cos(y))$

$\frac{p^2}{2}=18*\frac{cos^4(y)}{4}+C_0$

$p^2=9cos^4(y)+C$

$(y')^2=9cos^4(y)+C$

используем $y'(0)=3$

$9*1^4+C=3$

$(y')^2=9cos^4(y)-6$

$y'=\pm \sqrt{9cos^4(y)-6}$

$\frac{dy}{\sqrt{9cos^4(y)-6}}=\pm dx$

$\int\limits {\frac{dy}{\sqrt{9cos^4(y)-6}}} \, dx=\pm x+C$