Question

Помогите решить задачу по математике!)

Answer 1

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

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