Q.Solve the following differential equation: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0; y = 2$ when $x = 1$

Homogeneous (every term degree 2). Substitute $y=vx$.

Standard form: $2x^2\frac{dy}{dx} = 2xy + y^2 \implies \frac{dy}{dx} = \frac{y}{x} + \frac{1}{2}\left(\frac{y}{x}\right)^2$.

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

Separate and integrate: $\frac{2}{v^2}\,dv = \frac{dx}{x}$ gives $-\frac{2}{v} = \log|x| + C$. …

Spotting the type

Rewrite $2xy + y^2 - 2x^2\frac{dy}{dx}=0$ as

The right side depends only on $y/x$, so the equation is homogeneous.

Substitute $y=vx$

With $\frac{dy}{dx}=v+x\frac{dv}{dx}$,

Separate and integrate

Return to $x,y$

With $v=\frac{y}{x}$, $\frac{2}{v}=\frac{2x}{y}$, so …

Method: $y=vx$ for a homogeneous polynomial equation

Use this when $M\,dx+N\,dy=0$ (or an equivalent form) has every term of the same total degree — the sign that only the ratio $y/x$ matters.

Steps

Step 1: Get $\dfrac{dy}{dx}$ alone and check the degree.

Solve for $\dfrac{dy}{dx}$. If numerator and denominator are homogeneous of the same degree, dividing top and bottom by the leading power of $x$ turns the right side into a function of $y/x$.

Step 2: Substitute $y=vx$, $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Rewrite the equation entirely in $v$ and $x$.

Step 3: Subtract $v$ and separate variables. …

Common Mistakes

Mistake 1: Not reducing the right side to a function of $y/x$ first.

Why it's wrong: without writing $\dfrac{2xy+y^2}{2x^2}=\dfrac{y}{x}+\tfrac12\left(\tfrac{y}{x}\right)^2$, the substitution $y=vx$ won't cleanly cancel. Correct approach: divide through so only $y/x$ appears, then substitute.

Mistake 2: Cancelling $v$ incorrectly.

Why it's wrong: after substitution $v+x\dfrac{dv}{dx}=v+\tfrac{v^2}{2}$, only the plain $v$ cancels, leaving $x\dfrac{dv}{dx}=\tfrac{v^2}{2}$. Cancelling more (or the $v^2/2$) destroys the equation. Correct approach: subtract exactly the matching $v$. …