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# Solve: $x^2\Large \frac{dy}{dx}\normalsize =x^2+xy+y^2$

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• A linear differential equation of the form.$\large\frac{dy}{dx}$$=f(x,y) is a linear homogenous differential equation and it can be solved by substituting y=vx and \large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}.$
• $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg)+c Step 1: Given x^2 \large\frac{dy}{dx}$$=x^2+xy+y^2$
Divide throughout by $x^2$
$\large\frac{dy}{dx}=1+\frac{y}{x}+\frac{y^2}{x^2}$
Since this is a homologous differential equation.We can solve this equation by substituting $y=vx$ and $\large\frac{dy}{dx}$$=v+x\large\frac{dv}{dx} v+x \large\frac{dv}{dx}$$=1+\large\frac{vx}{x}+\frac{v^2x^2}{x^2}$
$=1+v+v^2$
$=>x \large\frac{dv}{dx}$$=1+v+v^2-v =>x \large\frac{dv}{dx}$$=1+v^2$
Step 2:
Now seperating the variables we get,
$\large\frac{dv}{1+v^2}= \frac{dx}{x}$
Now integrating on both sides we get,
$\large \int \frac{dv}{1+v^2}= \int \frac{dx}{x}$
$\large\frac{dv}{1+v^2}$ is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg) \large \int \frac{dv}{1+v^2}$$= \tan ^{-1}(v)$
Therefore $\tan ^{-1}(v)=\log |x|+c$
Substituting for $v=\large\frac{y}{x}$ we get,
$\tan ^{-1}\bigg(\large\frac{y}{x}\bigg)$$=\log(x)+c$