Q.Solve the following differential equation: $\left[x \sin^2 \left(\frac{y}{x}\right) - y\right] dx + x dy = 0; y = \frac{\pi}{4}$ when $x = 1$

Homogeneous (every term degree 1). Write the slope as a function of $y/x$ and put $y=vx$.

Standard form: $\frac{dy}{dx} = \frac{y - x\sin^2(y/x)}{x} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right)$.

Substitute $y=vx$, $\frac{dy}{dx}=v+x\frac{dv}{dx}$:

$v + x\frac{dv}{dx} = v - \sin^2 v \implies x\frac{dv}{dx} = -\sin^2 v.$ …

Spotting the type

In $[x\sin^2(y/x)-y]\,dx + x\,dy=0$ every term is degree 1 (the sine's argument $y/x$ is degree 0), so it is homogeneous and only $y/x$ matters.

Set up

Substitute $y=vx$

With $\frac{dy}{dx}=v+x\frac{dv}{dx}$ and $\sin^2(y/x)=\sin^2 v$,

Separate and integrate

$\frac{dv}{\sin^2 v} = -\frac{dx}{x} \implies \int\csc^2 v\,dv = -\int\frac{dx}{x}.$ …

Method: Homogeneous initial-value problem with $\sin^2(y/x)$ ($\csc^2$ integral)

Use this for $\big[x\sin^2\frac{y}{x}-y\big]dx+x\,dy=0$ with a condition; $y=vx$ yields a $\dfrac{dv}{\sin^2 v}$ (cosecant-squared) integral.

Steps

Step 1: Solve for $\dfrac{dy}{dx}$ and substitute $y=vx$

Rearrange to $\dfrac{dy}{dx}=v-\sin^2 v$, then set it equal to $v+x\dfrac{dv}{dx}$.

Step 2: Cancel and separate

The $v$ cancels: $x\dfrac{dv}{dx}=-\sin^2 v$, i.e. $\csc^2 v\,dv=-\dfrac{dx}{x}$.

Step 3: Integrate and apply the condition …

Common Mistakes

Mistake 1: Forgetting $\int\csc^2 v\,dv=-\cot v$

Why it's wrong: a wrong antiderivative here breaks the solution. Correct approach: recognise the $\sin^{-2}$ integrand and integrate to $-\cot v$.

Mistake 2: Sign error giving $x\frac{dv}{dx}=+\sin^2 v$

Why it's wrong: the correct reduction is $x\frac{dv}{dx}=-\sin^2 v$; the sign flips the constant. Correct approach: isolate carefully after cancelling $v$. …