Question
The general solution of the $D E x^2 \frac{d y}{d x}=x^2+x y+y^2$ is
✓
Answer
(a) $\tan ^{-1} \frac{y}{x}=\log x+C$
Explanation: We have,
$x^2 \frac{d y}{d z}=x^2+x y+y^2$
$\frac{d y}{d x}=1+\frac{y}{x}+\frac{y^2}{x^2}$Let $y = vx$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$1+v+v^2=v+x \frac{d v}{d x}$
$1+v^2=x \frac{d v}{d x}$
$\frac{d x}{x}=\frac{d v}{v^2+1}$
On integrating on both sides, we obtain
$\log x=\tan ^{-1} v+C$
$\tan ^{-1} \frac{y}{x}=\log x+c$
Explanation: We have,
$x^2 \frac{d y}{d z}=x^2+x y+y^2$
$\frac{d y}{d x}=1+\frac{y}{x}+\frac{y^2}{x^2}$Let $y = vx$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$1+v+v^2=v+x \frac{d v}{d x}$
$1+v^2=x \frac{d v}{d x}$
$\frac{d x}{x}=\frac{d v}{v^2+1}$
On integrating on both sides, we obtain
$\log x=\tan ^{-1} v+C$
$\tan ^{-1} \frac{y}{x}=\log x+c$
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