# Homogeneous Differential Equation

A differential equation of the form f(x,y)dy = g(x,y)dx is said to be **homogeneous differential equation** if the degree of f(x,y) and g(x, y) is same. A function of form F(x,y) which can be written in the form k^{n }F(x,y) is said to be a homogeneous function of degree n, for k≠0. Hence, f and g are the homogeneous functions of the same degree of x and y. Here, the change of variable y = ux directs to an equation of the form;

dx/x = h(u) du

which could be easily integrated.

Contrarily, a differential equation is homogeneous if it is a similar function of the anonymous function and its derivatives. For linear differential equations, there are no constant terms. The solutions of any linear ordinary differential equation of any degree or order may be calculated by integration from the solution of the homogeneous equation achieved by eliminating the constant term.

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(x/y)

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)=vF_{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(vx/vy)=v^{0}sin(vx/vy)=v^{0}F_{3}(x,y)

F_{4}(vx,vy)=sin(vx)+cos(vy)≠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.

## Steps to Solve Homogeneous Differential Equation

You must have learned to solve the differential equations in previous sections. 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=y/x

### Solved Example

**Solve dy/dx = (x-y)/(x+y)**

Solution: Given, dy/dx = (x-y)/(x+y)

Divide the RHS by x

(x/x-y/x)/(x/x+y/x) = (1-y/x)/(1+y/x)

Now, we can write;

dy/dx = (1-y/x)/(1+y/x)

If y = vx and dy/dx = v + xdv/dx

Then,

v + xdv/dx = (1-v)/(1+v)

Subtracting v from both the sides;

xdv/dx = (1-v)/(1+v) – v

xdv/dx = [(1-v)/(1+v)] -[(v+v^{2})/(1+v)]

xdv/dx = (1-2v-v^{2}))/(1+v)

Now we can use the separation of variables method;

(1+v)/(1-2v-v^{2}) dv = (1/x) dx

Integrating both the sides;

∫(1+v)/(1-2v-v^{2}) dv = ∫(1/x) dx

-1/2 ln(1-2v-v^{2}) = ln(x) + C

Put C=ln(k)

-1/2 ln(1-2v-v^{2}) = ln(x) + ln(k)

(1-2v-v^{2})^{-1/2} = kx

or we can write;

1-2v-v^{2} = 1/k^{2}x^{2}

Again, putting v = y/x;

1-2(y/x)-(y/x)^{2} = 1/k^{2}x^{2}

Eliminating x^{2} term from denominator on both the sides, we get;

x^{2}-2xy-y^{2} = 1/k^{2}

or

y^{2}+2xy-x^{2 }= -1/k^{2}

Now, put -/k^{2} = c

Adding 2x^{2} on both the sides;

y^{2}+2xy+x^{2} = c+2x^{2}

Now factoring the above equation, we get;

(y+x)^{2} = 2x^{2}+c

y+x=√(2x^{2})+c

Or y = ±√(2x^{2}+c) − x

This is the solution for the given equation.

### Nonhomogeneous Differential Equation

A linear nonhomogeneous differential equation of second order is represented by;

y”+p(t)y’+q(t)y = g(t)

where g(t) is a non-zero function.

The associated homogeneous equation is;

y”+p(t)y’+q(t)y = 0

which is also known as complementary equation.

### Examples of Homogeneous Equations

Q.1: 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.
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. |