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.
