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  • Prove that x^2 – y^2 = c (x^2 + y^2 ) 2 is the general solution of differential equation (x^3 – 3x y^2 ) dx = (y^3 – 3x^2 y) dy, where c is a parameter.

Q4. Prove that $x^2-y^2=c\left(x^2+y^2\right)^2$ is the general solution of differential equation $\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y$, where c is a parameter.

Answers (1)

best_answer

Given,
$\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y$
$\Longrightarrow \frac{d y}{d x}=\frac{\left(x^3-3 x y^2\right)}{\left(y^3-3 x^2 y\right)}$
Now, let $\mathrm{y}=\mathrm{vx}$
$\Longrightarrow \frac{d y}{d x}=\frac{d(v x)}{d x}=v+x \frac{d v}{d x}$
Substituting the values of $y$ and $y^{\prime}$ in the equation,

$v+x \frac{d v}{d x}=\frac{\left(x^3-3 x(v x)^2\right)}{\left((v x)^3-3 x^2(v x)\right)}$
$\Longrightarrow v+x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v} $
$\Longrightarrow x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}-v=\frac{1-v^4}{v^3-3 v}$
$\Longrightarrow\left(\frac{v^3-3 v}{1-v^4}\right) d v=\frac{d x}{x}$
Integrating both sides we get,
$\int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=\log x+\log C^{\prime}$

$\mathrm{Now}_{,} \int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=\int \frac{v^3}{1-v^4} d v-3 \int \frac{v d v}{1-v^4}$
$\Rightarrow \int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=I_1-3 I_2$, where $I_1=\int \frac{v^3}{1-v^4} d v$ and $I_2=\int \frac{v d v}{1-v^4}$
Let $1-v^4=\mathrm{t}$
$\frac{d}{d v}\left(1-v^4\right)=\frac{d t}{d v}$
$\Longrightarrow-4 v^3=\frac{d t}{d v}$
$\Longrightarrow v^3 d v=-\frac{d t}{4}$

Now,

$\mathrm{I}_1=\int-\frac{\mathrm{dt}}{4}=-\frac{1}{4} \log \mathrm{t}=-\frac{1}{4} \log \left(1-\mathrm{v}^4\right)$

and 
$I_2=\int \frac{v d v}{1-v^4}=\int \frac{v d v}{1-\left(v^2\right)^2}$
Let $v^2=p$
$\Rightarrow \frac{d}{d v}\left(v^2\right)=\frac{d p}{d v}$
$\Rightarrow 2 v=\frac{d p}{d v}$

$\Rightarrow \mathrm{vdv}=\frac{\mathrm{dp}}{2}$
$\therefore \mathrm{I}_2=\frac{1}{2} \int \frac{\mathrm{dp}}{1-\mathrm{p}^2}=\frac{1}{2 \times 2} \log \left|\frac{1+\mathrm{p}}{1-\mathrm{p}}\right|=\frac{1}{4}\left|\frac{1+\mathrm{v}^2}{1-\mathrm{v}}\right|$
Now, substituting the values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in the above equation, we get,
$\int\left(\frac{\mathrm{v}^3-3 \mathrm{y}}{1-\mathrm{v}^4}\right) \mathrm{dv}=-\frac{1}{4} \log \left(1-\mathrm{v}^4\right)-\frac{3}{4} \log \left|\frac{1+\mathrm{v}^2}{1-\mathrm{v}^2}\right|$
Thus,

$-\frac{1}{4} \log \left(1-\mathrm{v}^4\right)-\frac{3}{4} \log \left|\frac{1+\mathrm{v}^2}{1-\mathrm{v}^2}\right|=\log \mathrm{x}+\log \mathrm{C}^{\prime}$
$\Rightarrow-\frac{1}{4} \log \left[\left(1-\mathrm{v}^4\right)\left(\frac{1+\mathrm{v}^2}{1-\mathrm{v}^2}\right)^3\right]=\log \mathrm{C}^{\prime} \mathrm{x}$
$\Rightarrow \frac{\left(1+\mathrm{v}^2\right)^4}{\left(1-\mathrm{v}^2\right)^2}=\left(\mathrm{C}^{\prime} \mathrm{x}\right)^{-4}$

$\Rightarrow \frac{\left(1+\frac{\mathrm{y}^2}{\mathrm{x}^2}\right)^4}{\left(1-\frac{\mathrm{y}^2}{\mathrm{x}^2}\right)^2}=\frac{1}{\mathrm{C}^{\prime 4} \mathrm{x}^4}$
$\left(x^2-y^2\right)^2=C^{\prime 4}\left(x^2+y^2\right)^4$
$\Longrightarrow\left(x^2-y^2\right)=C^{\prime 2}\left(x^2+y^2\right)^2$
$\Longrightarrow\left(x^2-y^2\right)=K\left(x^2+y^2\right)^2, \text { where } K=C^{\prime 2}$

Posted by

HARSH KANKARIA

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