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 kⁿ 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,

F₁(x,y)=2x−8y

F₂(x,y)=x²+8xy+9y²

F₃(x,y) = sin(x/y)

F₄(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

F₁(vx,vy)=2(vx)−8(vy)=v(2x−8y)=vF₁(x,y)

F₂(vx,vy) = v²x² + 8(vx)(vy) + 9v²y² = v²(x²+8xy+9y²) = v²F₂(x,y)

F₃(vx,vy)=sin(vx/vy)=v⁰sin(vx/vy)=v⁰F₃(x,y)

F₄(vx,vy)=sin(vx)+cos(vy)≠vⁿF₄(x,y)

Hence, functions F₁, F₂, F₃ can be written in the form vⁿF(x,y), whereas F₄ cannot be written. Thus first three are homogeneous functions and the last function is not homogeneous.

Also, read:

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

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

Step 2 – Differentiating, we get,

Step 3 – Separating the variables, we get

Step 4 – Integrating both side of equation, we have

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-y)/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+v²)/(1+v)]

xdv/dx = (1-2v-v²))/(1+v)

Now we can use the separation of variables method;

(1+v)/(1-2v-v²) dv = (1/x) dx

Integrating both the sides;

∫(1+v)/(1-2v-v²) dv = ∫(1/x) dx

-1/2 ln(1-2v-v²) = ln(x) + C

Put C=ln(k)

-1/2 ln(1-2v-v²) = ln(x) + ln(k)

(1-2v-v²)^-1/2 = kx

or we can write;

1-2v-v² = 1/k²x²

Again, putting v = y/x;

1-2(y/x)-(y/x)² = 1/k²x²

Eliminating x² term from denominator on both the sides, we get;

x²-2xy-y² = 1/k²

or

y²+2xy-x² = -1/k²

Now, put -/k² = c

Adding 2x² on both the sides;

y²+2xy+x² = c+2x²

Now factoring the above equation, we get;

(y+x)² = 2x²+c

y+x=√(2x²)+c

Or y = ±√(2x²+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.

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Examples of Homogeneous Equations