Homogeneous Differential Equations

Solve:

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

Solve:

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

If $x\frac{dy}{dx}=y\left(\log y-\log x+1\right)$, then the solution of the equation is -

Which one of the following is/are homogeneous function(s)?

Let $f\left(x,y,c_1\right)=0$ and $f\left(x,y,c_2\right)=0$ define two integral curves of a homogeneous first order differential equation. If $P_1$ and $P_2$ are respectively the points of intersection of these curves with an arbitrary line, $y=mx$ then prove that the slopes of these two curves at $P_1$ and $P_2$ are equal.

Solve: $x^3\left(\frac{dy}{dx}\right)=y^3+y^2\sqrt{y^2-x^2}$.

Show that the curve such that the distance between the origin and the tangent at an arbitrary point is equal to the distance between the origin and the normal at the same point,$\sqrt{x^2+y^2}=c{e}^{\pm {\tan }^{-1}\frac{y}{x}}$

Use the substitution $y^2=a-x$ to reduce the equation $y^3·\frac{dy}{dx}+x+y^2=0$ to homogeneous form and hence solve it. $($where $a$ is variable$)$

The light rays emanating from a point source situated at origin when reflected from the mirror of a search light are reflected as beam parallel to the $x$-axis. Show that the surface is parabolic, by first forming the differential equation and then solving it.

Find the equation of a curve such that the projection of its ordinate upon the normal is equal to its. abscissa.

Solve : $\frac{dy}{dx}=\frac{2{\left(y+2\right)}^2}{{\left(x+y-1\right)}^2}$

Solve: $\frac{dy}{dx}=\frac{x+2y-3}{2x+y-3}$

Solve$:(x-y) dy=(x+y+1) dx$

Solve: $\frac{dy}{dx}=\frac{y-x+1}{y+x+5}$

A curve passing through the point $\left(1,1\right)$ has the property that the perpendicular distance of the origin from the normal at any point $P$ of the curve is equal to the distance of $P$ from the $x$-axis. Determine the equation of the curve.

Which of the following functions are not homogeneous?

Which of the following functions are homogeneous?

Solve:

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

A curve passes through the point $\left(1,\frac{\mathrm{\pi }}{4}\right)$ and its slope at any point is given by $\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$. Then, the curve has the equation