Question

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

Answer 1

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