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

Homogeneous. Divide by $x^2$: with $v=y/x$,

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

Separate and use partial fractions $\frac{1}{1-2v^2}=\frac12\!\left(\frac{1}{1-\sqrt2\,v}+\frac{1}{1+\sqrt2\,v}\right)$: …

1. Standard form

Every term is degree $2$, so the right side is a function of $v=y/x$ alone — the equation is homogeneous.

2. Substitute $y=vx$

3. Separate

4. Integrate the $v$-side by partial fractions

Since $1-2v^2=(1-\sqrt2\,v)(1+\sqrt2\,v)$,

so

Hence

5. Back-substitute $v=y/x$ …

Method: Homogeneous equation leading to a $\frac{1}{1-2v^2}$ integral

Use this for $x^2\dfrac{dy}{dx}=x^2-2y^2+xy$ and similar homogeneous equations where, after $y=vx$, the $v$-integral is a standard $\dfrac{1}{a^2-b^2v^2}$ (partial-fraction / log) form.

Steps

Step 1: Substitute $y=vx$

Divide by $x^2$ to get $\dfrac{dy}{dx}=1-2v^2+v$, then set $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Step 2: Cancel and separate

The $v$ terms cancel, leaving $x\dfrac{dv}{dx}=1-2v^2$, i.e. $\dfrac{dv}{1-2v^2}=\dfrac{dx}{x}$.

Step 3: Integrate with the standard log form …

Common Mistakes

Mistake 1: Not dividing through by $x^2$ first

Why it's wrong: you must express the right side in $v=\frac{y}{x}$ (i.e. $1-2v^2+v$) before substituting; skipping it hides the homogeneity. Correct approach: divide by $x^2$ then set $y=vx$.

Mistake 2: Misapplying the $\int\dfrac{dv}{1-2v^2}$ standard form …