Question

Solve the differential equation \[ x\frac{dy}{dx}=y-x\sin^2\left(\frac{y}{x}\right), \quad \text{given that } y(1)=\frac{\pi}{6}. \]

Hint:

When a differential equation contains the ratio \(y/x\), use the substitution \(v=\frac{y}{x}\). This converts the equation into a separable form.

Answer 1

Step 1: Identify the type of differential equation.
The given differential equation is \[ x\frac{dy}{dx}=y-x\sin^2\left(\frac{y}{x}\right) \] Divide both sides by \(x\): \[ \frac{dy}{dx}=\frac{y}{x}-\sin^2\left(\frac{y}{x}\right) \] Since the equation contains the ratio \(\frac{y}{x}\), it is a homogeneous differential equation.
Step 2: Use substitution.
Let \[ v=\frac{y}{x} \] Thus \[ y=vx \] Differentiate both sides: \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Step 3: Substitute into the equation.
\[ v+x\frac{dv}{dx}=v-\sin^2 v \] Cancel \(v\) from both sides: \[ x\frac{dv}{dx}=-\sin^2 v \] Step 4: Separate variables.
\[ \frac{dv}{\sin^2 v}=-\frac{dx}{x} \] Step 5: Integrate both sides.
Recall the identity \[ \frac{1}{\sin^2 v}=\csc^2 v \] Thus \[ \int \csc^2 v\,dv=-\int \frac{dx}{x} \] \[ -\cot v=-\ln|x|+C \] Multiply by \(-1\): \[ \cot v=\ln|x|+C \] Step 6: Substitute back $v=\frac{y{x}$.}
\[ \cot\left(\frac{y}{x}\right)=\ln|x|+C \] Step 7: Use the initial condition.
Given \[ y(1)=\frac{\pi}{6} \] Thus \[ \frac{y}{x}=\frac{\pi}{6} \] Substitute \(x=1\): \[ \cot\left(\frac{\pi}{6}\right)=\ln1+C \] \[ \sqrt3=0+C \] \[ C=\sqrt3 \] Step 8: Write the final solution.
\[ \cot\left(\frac{y}{x}\right)=\ln|x|+\sqrt3 \] Final Answer:
\[ \boxed{\cot\left(\frac{y}{x}\right)=\ln|x|+\sqrt3} \]