Q.Solve the following differential equation: $x^2 dy + (xy + y^2) dx = 0; y = 1$ when $x = 1$

The equation is homogeneous — every term has degree 2. Write the slope as a function of $y/x$ and substitute $y=vx$.

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

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

Separate and integrate using $\frac{1}{v(v+2)}=\frac12\left(\frac1v-\frac1{v+2}\right)$:

$\frac12\log\left|\frac{v}{v+2}\right| = -\log|x| + C \implies \frac{v}{v+2} = \frac{K}{x^2}.$ …

Spotting the type

Every term of $x^2\,dy + (xy+y^2)\,dx=0$ has total degree 2, so this is a homogeneous equation — its slope depends only on the ratio $y/x$. That is exactly what the substitution $y=vx$ is built for.

Set up

Solve for the derivative:

Substitute $y=vx$

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

Separate and integrate

Using partial fractions $\frac{1}{v(v+2)} = \frac12\left(\frac1v - \frac1{v+2}\right)$,

$\frac12\log\left|\frac{v}{v+2}\right| = -\log|x| + C \implies \frac{v}{v+2} = \frac{K}{x^2}.$ …

Method: Homogeneous initial-value problem via partial fractions in $v$

Use this for $x^2\,dy+(xy+y^2)\,dx=0$ with a condition; after $y=vx$ the $v$-integral factors and needs partial fractions, then the point gives the particular solution.

Steps

Step 1: Substitute $y=vx$

Rewrite as $\dfrac{dy}{dx}=-(v+v^2)$ (with $v=\dfrac{y}{x}$), then set it equal to $v+x\dfrac{dv}{dx}$.

Step 2: Separate and split by partial fractions

You get $\dfrac{dv}{v(v+2)}=-\dfrac{dx}{x}$; decompose $\dfrac{1}{v(v+2)}=\dfrac12\!\left(\dfrac1v-\dfrac{1}{v+2}\right)$. …

Common Mistakes

Mistake 1: Not using partial fractions on $\dfrac{1}{v(v+2)}$

Why it's wrong: it must split into $\frac12(\frac1v-\frac{1}{v+2})$; integrating it whole is impossible. Correct approach: decompose first.

Mistake 2: Sign error in $x\dfrac{dv}{dx}=-v(v+2)$

Why it's wrong: the minus sign controls the direction of the logs; losing it gives the wrong particular solution. Correct approach: carry the minus from $\frac{dy}{dx}=-(v+v^2)$. …