Дифференциальные уравнения. Помогите с решением 1,2 и 3 задания

Question 1

Question 2

$xy''+y'=4x^3\\y'=z;y''=z'\\xz'+z=4x^3\\z=uv;z'=u'v+v'u\\xu'v+xv'u+uv=4x^3\\xu'v+u(xv'+v)=4x^3\\\begin{cases}xv'+v=0\\u'v=4x^2\end{cases}\\\frac{xdv}{dx}+v=0|*\frac{dx}{xv}\\\frac{dv}{v}=-\frac{dx}{x}\\\int\frac{dv}{v}=-\int\frac{dx}{x}\\ln|v|=-ln|x|\\v=\frac{1}{x}\\\frac{du}{xdx}=4x^2\\du=4x^3dx\\\int du=4\int x^3dx\\u=x^4+C_1\\z=y'=x^3+\frac{C_1}{x}\\y=\frac{x^4}{4}+C_1ln|x|+C_2\\y'(1)=2\\2=1+C_1\\C_1=1\\y(1)=\frac{1}{4}\\\frac{1}{4}=\frac{1}{4}+C_2\\C_2=0\\y=\frac{x^4}{4}+ln|x|$
----------
$xy''-y'=x^2cosx\\y'=z;y''=z'\\xz'-z=x^2cosx\\z=uv;z'=u'v+v'u\\xu'v+xv'u-uv=x^2cosx\\xu'v+u(xv'-v)=x^2cosx\\\begin{cases}xv'-v=0\\u'v=xcosx\end{cases}\\\frac{xdv}{dx}-v=0|*\frac{dx}{vx}\\\frac{dv}{v}=\frac{dx}{x}\\\int\frac{dv}{v}=\int\frac{dx}{x}\\ln|v|=ln|x|\\v=x\\\frac{du}{dx}=cosx\\du=cosxdx\\\int du=\int cosxdx\\u=sinx+C_1\\z=y'=xsinx+C_1x\\y=-xcosx+sinx+\frac{C_1x^2}{2}+C_2\\y'(\frac{\pi}{2})=\frac{\pi}{2}\\\frac{\pi}{2}=\frac{\pi}{2}+\frac{\pi}{2}C_1\\C_1=0\\y(\frac{\pi}{2})=1\\1=1+C_2\\C_2=0\\$
$y=-xcosx+sinx$
----------
$x^3y''=4lnx\\y''=\frac{4lnx}{x^3}\\y'=\int\frac{4lnx}{x^3}dx=-2\frac{lnx}{x^2}+2\int\frac{dx}{x^3}=-\frac{2lnx+1}{x^2}+C_1\\y=\int(-\frac{2lnx+1}{x^2}+C_1)dx=\frac{2lnx+3}{x}+C_1x+C_2\\y'(1)=0\\0=-1+C_1\\C_1=1\\y(1)=4\\4=3+1+C_2\\C_2=0\\y=\frac{2lnx+3}{x}+x$