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Methods of Solving Differential Equations> Homogeneous Differential Equations

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Estimated time: 9 minutes
CBSE: Class 12
Maharashtra State Board: Class 12

Definition: Homogeneous Differential Equations

A differential equation of the form \[\frac{dy}{dx}=\frac{f_{1}(x,y)}{f_{2}(x,y)},\] where f1(x, y) ) and f2(x, y)  are homogeneous functions of x and y of the same degree, is called a homogeneous differential equation.

CBSE: Class 12

Method

  • Rewrite the differential equation in a simplified form.

  • Check whether the right-hand side depends only on \[\frac{y}{x}\] or \[\frac{x}{y}\].

  • Choose substitution: y = vx or x = vy.

  • Differentiate the substitution correctly using the product rule.

  • Reduce the equation to a separable form.

  • Integrate both sides.

  • Replace v by \[\frac{y}{x}\] or \[\frac{x}{y}\] to obtain the final answer.

CBSE: Class 12

Example 1

Example 11 – Stepwise

Question: Show that \[x \cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x\] is homogeneous and solve it.

Step 1: Write in standard form

\[x \cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \implies \frac{dy}{dx} = \frac{y \cos\left(\frac{y}{x}\right) + x}{x \cos\left(\frac{y}{x}\right)}\]

So it is of the form \[\frac{dy}{dx} = F(x, y)\].

Step 2: Check homogeneity

Let

\[F(x, y) = \frac{y \cos\left(\frac{y}{x}\right) + x}{x \cos\left(\frac{y}{x}\right)}\]

Then

\[F(\lambda x, \lambda y) = \frac{\lambda y \cos\left(\frac{\lambda y}{\lambda x}\right) + \lambda x}{\lambda x \cos\left(\frac{\lambda y}{\lambda x}\right)} = \frac{\lambda\left(y \cos\left(\frac{y}{x}\right) + x\right)}{\lambda\left(x \cos\left(\frac{y}{x}\right)\right)} = \frac{y \cos\left(\frac{y}{x}\right) + x}{x \cos\left(\frac{y}{x}\right)} = F(x, y)\]
 
So \[F(x, y)\] is homogeneous of degree 0; the equation is homogeneous.

Step 3: Substitution

Put

\[y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\]

Substitute into the equation:

\[v + x\frac{dv}{dx} = \frac{v \cos v + 1}{\cos v}\]

So

\[x\frac{dv}{dx} = \frac{v \cos v + 1}{\cos v} - v = \frac{1}{\cos v}\]

Hence

\[\cos v \, dv = \frac{dx}{x}\]

Step 4: Integration

\[\int \cos v \, dv = \int \frac{dx}{x}\]
\[\sin v = \ln|x| + \ln|C|\]
\[\implies \sin v = \ln|Cx|\]

Step 5: Back substitution

Replace \[v = \frac{y}{x}\]:

\[\sin\left(\frac{y}{x}\right) = \ln|Cx|\]
CBSE: Class 12
Maharashtra State Board: Class 12

Key Points: Homogeneous Differential Equations

  • Check homogeneity first.

  • Differentiate substitution carefully.

  • Convert to separable form.

  • Back-substitute to original variables.

Video Tutorials

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