# Homogeneous Differential Equation

A function of the form $F(x,y)$ which can be written in the form $k^{n}F(x,y)$, is said to be homogeneous function of degree n, for $k \neq 0$.

Or

A differential equation of the form $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$ is said to be homogeneous differential equation if the degree of the numerator i.e. $f(x, y)$ and the denominator $g(x, y)$ is same.

Consider the following functions in x and y,

$F_{1} (x,y) = 2x- 8y$

$F_{2} (x,y) = x^{2}+ 8xy +9y^{2}$

$F_{3} (x,y) = \sin \left ( \frac{x}{y} \right )$

$F_{4} (x,y) = \sin x + \cos y$

If we replace x and y with vx and vy respectively, for non-zero value of v, we get

$F_{1} (vx,vy) = 2(vx)- 8(vy) = v(2x- 8y) =v \; F_{1} (x,y)$

$F_{2} (vx,vy) = v^{2}x^{2}+ 8(vx)(vy) +9v^{2}y^{2}=v^{2}(x^{2}+ 8xy +9y^{2}) = v^{2} F_{2} (x,y)$

$F_{3} (vx,vy) = \sin \left ( \frac{vx}{vy} \right ) = v^{0}\sin \left ( \frac{vx}{vy} \right )= v^{0}F_{3} (x,y)$

$F_{4} (vx,vy) = \sin (vx) + \cos (vy) \neq v^{n}F_{4}(x,y)$

Hence, functions $F_{1},F_{2},F_{3}$ can be written in the form $v^{n}F(x,y)$, whereas $F_{4}$ cannot be written. Thus first three are homogeneous functions and the last function is not homogeneous.

Solving Homogeneous Differential Equation-

To solve a homogeneous differential equation following steps are followed:-

Given differential equation of the type $\frac{\mathrm{d} y}{\mathrm{d} x} = F(x,y) = g\left ( \frac{y}{x} \right )$

Step 1- Substitute $y = vx$ in the given differential equation.

Step 2 – Differentiating, we get, $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Now substitute the value of and y in the given differential equation, we get

$v + x\frac{dv}{dx}= g(v)$

$\Rightarrow x\frac{dv}{dx} = g(v)- v$

Step 3 – Separating the variables, we get

$\frac{dv}{g(v)-v}=\frac{dx}{x}$

Step 4 – Integrating both side of equation, we have

$\int \frac{dv}{g(v)-v} dv =\int \frac{dx}{x} + C$

Step 5 – After integration we replace $v = \frac{y}{x}$

 Lets Work Out- Example: Find the equation of the curve passing through the point $(2, \frac{\pi}{3} )$ when the tangent at any point makes an angle $tan^{-1}(\frac{y}{x} – sin^2 \frac{y}{x})$. Solution: $\phi = tan^-1 (\frac{y}{x} – sin^2 \frac{y}{x})$ Or $\frac{dy}{dx} = tan\phi = \frac{y}{x} – sin^2 \frac{y}{x}$ Since this equation represents a differential equation of homogeneous type therefore we substitute $y = vx$ in the above equation. $\Rightarrow v + x\frac{dv}{dx} = v – sin^2 v$ $\Rightarrow x \frac{dv}{dx} = – sin^2 v$ $\Rightarrow \frac{dx}{x} = -cosec^2 v dv$ Now integrating both the sides w.r.t. to x and v respectively, we get $\int \frac{dx}{x} = \int -cosec^2 v dv$ $ln x = \frac{1}{tan v} + C$ …………………(i) Also as it passes through the point $(2, \frac{\pi}{3})$, for (x,y). We know that v = y/x, thus value of v = $\frac{\pi}{3} \div 2 = \frac{\pi}{6}$ So substituting the values of x and v in the equation (i), we get $ln 2 = \sqrt{3} + C$ $\Rightarrow C = ln 2 – \sqrt{3}$ Or $ln x = \frac{1}{tan v} + ln 2 – \sqrt{3}$ Or $ln x = \frac{1}{tan\frac{y}{x}} + ln 2 – \sqrt{3}$ This is the required solution. Example: Find the equation of the curve passing through the point (1,-2) when the tangent at any point is given by $\frac{y(x + y^3)}{x(y^3 – x)}$. Solution: The equation of tangent represents the slope of the curve i.e. This equation is homogeneous in nature. On cross-multiplication, we get- $(xy^3 – x^2)dy = (xy + y^4)dx$ Solving the equation, we get $x^{2}y^{3} \frac{\left( xdy – ydx \right )}{x^{2}} – x (xdy – ydx) = 0$ $\Rightarrow x^{2}y^{3} d\frac{y}{x} – xd(xy) = 0$ Dividing both the sides by $x^3y^2$ we get, $\frac{y}{x} d(\frac{y}{x}) – \frac{d(xy)}{x^2y^2} = 0$ Now integrating this equation with respect to $\frac{y}{x} and xy$ we have, $\int \frac{y}{x} d(\frac{y}{x}) = \int \frac{d(xy)}{x^2y^2}$ $\frac{1}{2} \left ( \frac{y}{x} \right )^{2} = – \frac{1}{xy} +C$ ………………. (1) Now substituting the value of the given point in the above equation, we have $\Rightarrow \frac{1}{2} \times 4 – \frac{1}{2} = C$ $\Rightarrow C = \frac{3}{2}$ Put this value of the constant C in equation (1) we get $\frac{1}{2} (\frac{y}{x})^2 + \frac{1}{xy} = \frac{3}{2}$< This is the required solution.