Помогите с дифференциаламиa) y'=(2y+1)×tg xb) xy'-y=x×tg(y/x)c) y'+y=x/y^2

Решите задачу:

$\displaystyle \frac{dy}{dx}=(2y+1)tgx|*\frac{dx}{2y+1}\\\frac{dy}{2y+1}=tgxdx\\\frac{1}{2}\int\frac{d(2y+1)}{2y+1}=-\int\frac{d(cosx)}{cosx}\\\frac{1}{2}ln|2y+1|=-ln|cosx|+C|*2\\ln|2y+1|=-2ln|cosx|+ln|C^*|\\2y+1=\frac{C^*}{cos^2x}\\y=\frac{C^*-cos^2x}{2cos^2x}$

$xy'-y=x*tg\frac{y}{x}\\y=tx;y'=t'x+t\\x(t'x+t)-tx=xtgt|:x\\t'x+t-t=tgt\\\frac{dt}{dx}x=tgt|*\frac{ctgtdx}{x}\\ctgtdt=\frac{dx}{x}\\\int\frac{d(sint)}{sint}=\int\frac{dx}{x}\\ln|sint|=ln|x|+ln|C|\\sint=Cx\\\frac{1}{x}sin\frac{y}{x}=C$

$y'+y=\frac{x}{y^2}|*y^2\\y^2y'+y^3=x\\z=y^3;z'=3y^2y';y^2y'=\frac{z'}{3}\\\frac{z'}{3}+z=x\\z'+3z=3x\\z=uv;z'=u'v+v'u\\u'v+v'u+3uv=3x\\u'v+u(v'+3v)=3x\\\begin{cases}v'+3v=0\\u'v=3x\end{cases}\\\frac{dv}{dx}+3v=0|*\frac{dx}{v}\\\frac{dv}{v}=-3dx\\\int\frac{dv}{v}=-3\int dx\\ln|v|=-3x\\v=e^{-3x}\\\frac{du}{dx}e^{-3x}=3x\\du=3xe^{3x}\\\int du=\int 3xe^{3x}\\u=xe^{3x}-\frac{e^{3x}}{3}+C\\z=y^3=x-\frac{1}{3}+Ce^{-3x}\\y=\sqrt[3]{x-\frac{1}{3}+Ce^{-3x}}$