Q.Solve the following differential equation: $(x^2 + xy) dy = (x^2 + y^2) dx$

Homogeneous (every term is degree 2). Put $y=vx$, $\frac{dy}{dx}=v+x\frac{dv}{dx}$ into $\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}=\frac{1+v^2}{1+v}$:

Separate and write $\frac{1+v}{1-v}=-1+\frac{2}{1-v}$:

With $v=\frac{y}{x}$ and $1-v=\frac{x-y}{x}$ this tidies to $(x-y)^2=Cx\,e^{-y/x}$.

1. Recognise homogeneity

Numerator and denominator are both degree $2$, so dividing through by $x^2$ makes the right side depend only on $v=y/x$:

2. Substitute $y=vx$

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

3. Reduce to separable form

So

4. Integrate

Write $\dfrac{1+v}{1-v}=-1+\dfrac{2}{1-v}$:

5. Return to $x,y$

With $v=\frac{y}{x}$ and $1-v=\frac{x-y}{x}$:

Collecting logarithms, $-\frac{y}{x}-2\log|x-y|+2\log|x|=\log|x|+C$, i.e. $\log\dfrac{|x|}{(x-y)^2}=\dfrac{y}{x}+C$. Exponentiating,

Check: differentiating $(x-y)^2=Cx\,e^{-y/x}$ implicitly and simplifying returns $y'=\dfrac{x^2+y^2}{x^2+xy}$, the original equation.

Method: Solving a homogeneous equation by $y=vx$

Use this when $\dfrac{dy}{dx}$ can be written purely in terms of $\dfrac{y}{x}$ — a homogeneous equation. The substitution $y=vx$ turns it into a separable equation in $v$ and $x$.

Steps

Step 1: Confirm homogeneity and substitute

Rewrite the right side as a function of $v=\dfrac{y}{x}$. Set $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Step 2: Separate $v$ from $x$

After substituting, the $v$ terms group on one side and $\dfrac{dx}{x}$ on the other:

Step 3: Integrate and back-substitute

Integrate both sides, then replace $v$ with $\dfrac{y}{x}$ to return to $x,y$.

Common Mistakes

Mistake 1: Forgetting $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ after $y=vx$

Why it's wrong: writing only $\frac{dy}{dx}=v$ omits the product-rule term and breaks the whole method. Correct approach: differentiate $y=vx$ properly to $v+x\frac{dv}{dx}$.

Mistake 2: Errors simplifying $\dfrac{1+v^2}{1+v}-v$

Why it's wrong: this must reduce to $\frac{1-v}{1+v}$ before separating; algebra slips give a wrong integral. Correct approach: combine over a common denominator carefully.

Mistake 3: Not back-substituting $v=\dfrac{y}{x}$

Why it's wrong: leaving the answer in $v$ hides the actual $x,y$ solution. Correct approach: replace $v$ to reach $(x-y)^2=Cxe^{-y/x}$.