Помогите решить задачу по математике!)

$yy'=2y-x \\y'=2- \frac{x}{y} \\ y'=2- \frac{1}{ \frac{y}{x} } \\ u=\frac{y}{x} \ y=ux \\ y'=u+xu'\\ u+xu'=2- \frac{1}{u} \\ xu'=2- \frac{1}{u} -u\\ x \frac{du}{dx} =\frac{2u-1-u^2}{u}\\ -\int \frac{udu}{u^2-2u+1} = \int \frac{dx}{x} \\ -\int \frac{(u-1)+1}{(u-1)^2} \, du = \int \frac{dx}{x} \\ - \int \frac{du}{u-1} - \int \frac{du}{(u-1)^2} = \int \frac{dx}{x} \\ -ln|u-1|+ \frac{1}{u-1} =ln|x|+C \\ \frac{1}{u-1} e^{ \frac{1}{u-1}}=Cx\\ \frac{1}{ \frac{y}{x} -1} e^{ \frac{1}{ \frac{y}{x} -1}}=Cx\\$
$\frac{x}{y-x} e^{ \frac{x}{y-x}}=Cx\\ e^{ \frac{x}{y-x}}=C(y-x)\\ y-x=Ce^{ \frac{x}{y-x}}$

$y'-2xy=2xe^{x^2} \\ y=uv \\ y'=u'v+uv' \\ u'v+uv'-2xuv=2xe^{x^2} \\ u'v+u(v'-2xv)=2xe^{x^2} \\ v'-2xv=0 \\ \frac{dv}{dx}-2xv=0 \\ \int \frac{dv}{v}=\int 2xdx \\ ln|v|=x^2+C \\ v=Ce^{x^2} \\ v=e^{x^2} \ (C=1) \\ u'e^{x^2}=2xe^{x^2} \\ u'=2x \\ u=x^2+C \\ y=uv=(x^2+C)e^{x^2}$
Ответ: $y=(x^2+C)e^{x^2}$