Homogeneous Differential Equations

If $y=y\left(x\right)$ is the solution of the differential equation, $x\frac{dy}{dx}=y\left({\log }_{\mathrm{e}}y-{\log }_{\mathrm{e}}x+1\right)$, when $y\left(1\right)=2,$ then $y\left(2\right)$ is equal to________________

Find the equation of a curve passing through the point $\left(1, \frac{\pi }{4}\right)$ and having differential equation $\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$.

The solution of differential equation $(1-xy+x^2{y}^2)dx=x^{2}dyis$ :

Solve:

$\frac{dy}{dx}=\frac{y}{x}+\sin \left(\frac{y}{x}\right)$

Solution of the differential equation $\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$ is

The solution of $\frac{dy}{dx}=\frac{2y+x+1}{2x+4y+3}$

Solve the differential equation $\left(x^2+y^2\right)dx-2xydy=0$

Solve: $y^{\text{'}}=\frac{6x^2-5xy-2y^2}{6x^2-8xy+y^2}$

Find the general solution of the equation

$\left[2x\sin \left(\frac{y}{x}\right)+2x\tan \left(\frac{y}{x}\right)-y\cos \left(\frac{y}{x}\right)-y{\sec }^2\left(\frac{y}{x}\right)\right]dx+\left[x\cos \left(\frac{y}{x}\right)+x{\sec }^2\left(\frac{y}{x}\right)\right]dy=0$

Solve the differential equation $x(x-y)dy=y(x+y)dx$

The general solution of $\frac{dy}{dx}=\frac{x+y+1}{y-x+1}$ is

Solve the following differential equation: $x dy-y dx=\sqrt{x^2+y^2}dx$.

Solve the following differential equation: $x^{2}y dx-\left(x^3+y^3\right)dy=0$.

If the slope of the tangent at $\left(x,y\right)$ to a curve passing through $\left(1,\frac{\pi }{4}\right)$ is given by $\frac{y}{x}-{\cos }^2 \left(\frac{y}{x}\right)$, then the equation of the curve is

Solution of the differential equation $x{e}^{\frac{y}{x}}-y\sin \left(\frac{y}{x}\right)+x\frac{dy}{dx}\sin \left(\frac{y}{x}\right)=0$, given $y\left(1\right)=0$ is

General solution of differential equation $\frac{dy}{dx}=\frac{y-x}{x+y}$ is

Solve  $x^{2}y dy+\left(x^3+x^{2}y-2x{y}^2-y^3\right)dx=0$

Solve the differential equation, $y\cos \frac{y}{x}\left(xdy-ydx\right)+x\sin \frac{y}{x}\left(xdy+ydx\right)=0,$ when $y\left(1\right)=\frac{\mathrm{\pi }}{2}$.

The solution $y=y\left(x\right)$ of the differential equation $\left(x^2+y^2\right)dy=xydx$ satisfies the conditions $y\left(1\right)=1$ and $y\left(x_0\right)=e,$ then value of $x_0$ is

The curve amongst the family of curves represented by the differential equation, $\left(x^2-y^2\right)dx+2xydy=0,$ which passes through $(1, 1),$ is