Answer to Question #106142 in Differential Equations for Yogesh

Question

Solve the differential equation dy/dx+xy=y^2e^(x^2/2)sinx

Answer 1

$y'+xy=y^2e^{x^2/2}sinx$

divide both sides by $y^2$

$y'/y^2+x/y=e^{x^2/2}sinx$

substitution: $t=1/y$

$xt-t'=e^{x^2/2}sinx$

let's solve first this equation: $xt-t'=0$

$\int dt/t=\int xdx$

$lnt-lnC=x^2/2$

$t=Ce^{x^2/2}$

then solution to initial equation will have form: $t(x)=C(x)e^{x^2/2}$

$t'(x)=C'(x)e^{x^2/2}+xC(x)e^{x^2/2}$

Let's put these to initial equation

$xC(x)e^{x^2/2}-C'(x)e^{x^2/2}-xC(x)e^{x^2/2}=e^{x^2/2}sinx$

$C'(x)=-sinx$

$C(x)=\int -sinxdx=C_1+cosx$

then $t(x)=(C_1+cosx)e^{x^2/2}$

$y=1/t=e^{-x^2/2}/(C_1+cosx)$