Помогите решить ДУ: 2^(x+y)+3^(x-2*y)*y'=0 y(0)=0

Решите задачу:

$2^{x+y}+3^{x-2y}y'=0\\2^x*2^y+\frac{3^x}{3^{2y}}\frac{dy}{dx}=0|*\frac{dx}{3^x*2^y}\\\frac{2}{3}^xdx+\frac{dy}{18^y}=0\\\frac{2}{3}^xdx=18^{-y}d(-y)\\\frac{\frac{2}{3}^x}{ln\frac{2}{3}}=\frac{18^{-y}}{ln18}+C\\\frac{\frac{2}{3}^x}{ln\frac{2}{3}}-\frac{18^{-y}}{ln18}=C$
$y(0)=0\\\frac{\frac{2}{3}^0}{ln\frac{2}{3}}-\frac{18^{0}}{ln18}=C\\\frac{1}{ln\frac{2}{3}}-\frac{1}{ln18}=C\\C=log_\frac{2}{3}e-log_{18}e$