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

1 Answer

  • 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 $x^2$
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}$
$=>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$
answered May 21, 2013 by meena.p