Solution of Homogeneous Differential Equation

Here you will learn how to find solution of homogeneous differential equation of first order first degree with examples.

Let’s begin –

Solution of Homogeneous Differential Equation

If a first degree first order differential equation is expressible in the form

\(dy\over dx\) = \(f(x, y)\over g(x, y)\)

where f(x, y) and g(x, y) are homogeneous function of the same degree, then it is called a homogeneous differential equation,

Such type of equation can be reduced to variable seperable form by the substitution y = vx. Process is shown in the algorithm below.

Algorithm :

1). Put the differential equation in the form

\(dy\over dx\) = \(f(x, y)\over g(x, y)\)

2). Put y = vx and \(dy\over dx\) = v + x\(dy\over dx\) in the equation in step 1 and cancel out x from the right hand side.

3). Shift v on RHS and seperate the variables v and x.

4). Integrate both sides to obtain the solution in terms of v and x.

5). Replace v by \(y\over x\) in the solution obtained in step 4 to obtain the solution in terms of x and y.

Example : Solve the differential equation \(x^2\) dy + y(x + y) dx = 0.

Solution : The given differential equation is

\(x^2\) dy + y(x + y) dx = 0

\(\implies\) \(x^2\) dy = -y(x + y) dx

\(\implies\) \(dy\over dx\) = -(\(xy + y^2\over x^2\))                      …………………(i)

Since each of the functions xy + \(y^2\) and \(x^2\) is a homogeneous function of degree 2.

Therefore, equation (i) is a homogeneous differential equation.

Putting y = vx and \(dy\over dx\) = v + x\(dv\over dx\) in (i), we get

v + x\(dv\over dx\) = -(\(vx^2 + v^2x^2\over x^2\))

v + x\(dv\over dx\) = -(\(v + v^2\))

x\(dv\over dx\) = -(\(2v + v^2\))

xdv = -(\(v^2 + 2v\))dx

By Seperating the variable,

\(1\over v^2 + 2v\) dv = \(-dx\over x\)

Integrating both sides.

\(\implies\) \(\int\) \(1\over v^2 + 2v\) dv = \(\int\) \(-1\over x\) dx

\(\int\) \(1\over v^2 + 2v + 1- 1\) dv = \(\int\) \(-1\over x\) dx

\(\int\) \(1\over (v + 1)^2 – 1^2\) dv = \(\int\) \(-1\over x\) dx

\(1\over 2\) \(log{{v+1 – 1}\over {v+1+1}}\) = -log x + log C

\(1\over 2\) \(log{{v}\over {v+2}}\) = -log x + log C

\(log{{v}\over {v+2}}\) + 2log x =  2log C

\(log|{vx^2\over v+2}|\) = log k

Put v = y/x

k = \(x^2y\over y + 2x\), which is the solution of differential equation.

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