Помогите решить пожалуйста срочно!!!!!!!!!!!

Решите задачу:

$\displaystyle\it xdx=ydy\\\int xdx=\int ydy\\\frac{x^2}{2}=\frac{y^2}{2}+C\\\\\frac{x^2}{2}-\frac{y^2}{2}=C\\\\y(2)=1;\\\\2-\frac{1}{2}=C\\\\C=\frac{3}{2}\\\\\frac{x^2}{2}-\frac{y^2}{2}=\frac{3}{2}$

$\displaystyle\it x^2dy-\frac{1}{2}y^3dx=0|*\frac{1}{x^2y^3}\\\frac{dy}{y^3}=\frac{dx}{2x^2}\\\int\frac{dy}{y^3}=\frac{1}{2}\int\frac{dx}{x^2}\\-\frac{1}{2y^2}=-\frac{1}{2x}+C\\-\frac{1}{2y^2}+\frac{1}{2x}=C\\y(-1)=1\\\\-\frac{1}{2}-\frac{1}{2}=C\\\\C=-1\\\\-\frac{1}{2y^2}+\frac{1}{2x}=-1$

$\displaystyle\it y'+\frac{y}{x}=\frac{3}{x}\\y=uv;y'=u'v+v'u\\u'v+u(v'+\frac{v}{x})=\frac{3}{x}\\\begin{cases}v'+\frac{v}{x}=0\\u'v=\frac{3}{x}\end{cases}\\\frac{dv}{dx}+\frac{v}{x}=0|*\frac{dx}{v}\\\frac{dv}{v}=-\frac{dx}{x}\\\int\frac{dv}{v}=-\int\frac{dx}{x}\\ln|v|=-ln|x|\\v=\frac{1}{x}\\\frac{du}{xdx}=\frac{3}{x}|*xdx\\\\du=3dx\\\int du=3\int dx\\u=3x+C\\y=3+\frac{C}{x}$

$\displaystyle\it y''=x+1\\y'=\int(x+1)dx\\y'=\frac{x^2}{2}+x+C_1\\y=\int(\frac{x^2}{2}+x+C_1)dx=\frac{x^3}{6}+\frac{x^2}{2}+C_1x+C_2$

$y''+2y'+2y=0\\\lambda^2+2\lambda+2=0\\\lambda_{1,2}=-1^+_-i\\y=e^{-x}(C_1cosx+C_2sinx)\\y(0)=1\\1=C_1\\y'=-e^{-x}(C_1cosx+C_2sinx)+e^{-x}(-C_1sinx+C_2cosx)\\y'(0)=1\\1=-C_1+C_2\\C_2=2\\y=e^{-x}(cosx+2sinx)$