Chapter 3: First Order Homogeneous Differential Equations

First Order Homogeneous Differential Equations (Solved Examples)

A differential equation of the form

\[ \frac{dy}{dx}=f\left(\frac{y}{x}\right) \]

is called a homogeneous differential equation. We use the substitution

\[ y=vx \quad \text{or} \quad v=\frac{y}{x} \]

so that

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

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Example 1: Solve \( \frac{dy}{dx}=\frac{x+y}{x} \)

View Solution

\[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ \frac{dv}{dx}=\frac{1}{x} \] Integrate \[ v=\ln|x|+C \] Since \(v=\frac{y}{x}\) \[ \frac{y}{x}=\ln|x|+C \] \[ y=x(\ln|x|+C) \]

Example 2: Solve \( \frac{dy}{dx}=\frac{x+y}{x-y} \)

View Solution

Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=\frac{1+v}{1-v} \] \[ x\frac{dv}{dx}=\frac{1+v}{1-v}-v \] \[ x\frac{dv}{dx}=\frac{1+v^2}{1-v} \] Separate variables \[ \frac{1-v}{1+v^2}dv=\frac{dx}{x} \] Integrate \[ \tan^{-1}v-\frac12\ln(1+v^2)=\ln|x|+C \]

Example 3: Solve \( \frac{dy}{dx}=\frac{x-y}{x+y} \)

View Solution

Let \[ v=\frac{y}{x} \] \[ y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=\frac{1-v}{1+v} \] \[ x\frac{dv}{dx}=\frac{1-v}{1+v}-v \] Solve by separation of variables.

Example 4: Solve \( \frac{dy}{dx}=\frac{y}{x}+1 \)

View Solution

\[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ v=\ln|x|+C \] \[ y=x(\ln|x|+C) \]

Example 5: Solve \( \frac{dy}{dx}=\frac{x^2+y^2}{xy} \)

View Solution

\[ \frac{dy}{dx}=\frac{x}{y}+\frac{y}{x} \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=\frac{1}{v}+v \] \[ x\frac{dv}{dx}=\frac{1}{v} \] \[ v\,dv=\frac{dx}{x} \] Integrate \[ \frac{v^2}{2}=\ln|x|+C \] \[ \left(\frac{y}{x}\right)^2=2\ln|x|+C \]

Example 6: Solve \( \frac{dy}{dx}=\frac{x+y}{y} \)

View Solution

\[ \frac{dy}{dx}=\frac{x}{y}+1 \] Let \[ v=\frac{y}{x} \] Substitute \(y=vx\) and solve the resulting separable equation.

Example 7: Solve \( \frac{dy}{dx}=\frac{y-x}{x} \)

View Solution

\[ \frac{dy}{dx}=\frac{y}{x}-1 \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=v-1 \] \[ x\frac{dv}{dx}=-1 \] \[ v=-\ln|x|+C \] \[ \frac{y}{x}=-\ln|x|+C \]

Example 8: Solve \( \frac{dy}{dx}=\frac{x^2-y^2}{xy} \)

View Solution

\[ \frac{dy}{dx}=\frac{x}{y}-\frac{y}{x} \] Let \[ y=vx \] Substitute and solve the separable equation.

Example 9: Solve \( \frac{dy}{dx}=\left(\frac{y}{x}\right)^2 \)

View Solution

Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=v^2 \] \[ x\frac{dv}{dx}=v^2-v \] Separate variables \[ \frac{dv}{v(v-1)}=\frac{dx}{x} \] Integrate using partial fractions.

Example 10: Solve \( \frac{dy}{dx}=\frac{x^2+xy}{x^2} \)

View Solution

\[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ v=\ln|x|+C \] \[ y=x(\ln|x|+C) \]