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.

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Practise This Question

The general solution of dydxyx=y2x2 is