Solve the following problems:

$\frac{1}{x}\frac{dx}{dy}=\frac{ysin(x)}{y^2+1}$

$(\frac{1}{x}+2y^{2}x)dx+(2y{x}^2-cos(y))dy=0,y(1)=$

$(x^2+1)\frac{dy}{dx}+xy-x=0.$

$(2y^2+2y+4x^2)dx+(2xy+x)dy$

$\frac{dy}{dx}=\frac{y(ln(y)-ln(x)+1)}{x}$

$\frac{dy}{dx}=\sqrt[\phantom{2}]{x+y}-1$

$\frac{dy}{dx}=\frac{2y}{x}-x^2{y}^2$