Let y=y(x) be the solution of the differential equation (x^2-3y^2)dx+3xy~dy=0,y(1)=1.

Question

Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xy~dy=0,y(1)=1$. Then $6y^2(e)$ is equal to

Accepted Answer

Given,

$$\left( {{x^2} - 3{y^2}} \right)dx + 3xydy = 0$$

$$ \Rightarrow {x^2}dx - 3{y^2}dx + 3xydy = 0$$

$$ \Rightarrow {{{x^2}dx} \over {3xdx}} - {{3{y^2}dx} \over {3xdx}} + {{3ydy} \over {3xdx}} = 0$$

$$ \Rightarrow {x \over 3} - {{{y^2}} \over x} + y{{dy} \over {dx}} = 0$$

Let ${y^2} = t$

$$2y{{dy} \over {dx}} = {{dt} \over {dx}}$$

$$\therefore$$ $${1 \over 2}{{dt} \over {dx}} - {t \over x} + {x \over 3} = 0$$

$$ \Rightarrow {{dt} \over {dx}} - {{2t} \over x} =  - {{2x} \over 3}$$

This is linear Differential Equation.

$$
\text { I.F. }=e^{\int-\frac{2}{x} d x}=e^{-2 \ln |x|}=e^{\ln \frac{1}{x^2}}=\frac{1}{x^2}
$$

So, $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(\frac{-2 x}{3}\right) d x
$

$\Rightarrow \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+C$

When, $x=1, y=1$

$$
1=\frac{-2}{3} \ln (1)+C \Rightarrow C=1 $$

$$ \therefore \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+1
$$

At $x=e, \frac{y^2(e)}{e^2}=\frac{-2}{3}+1 $

$\Rightarrow y^2(e)=\frac{e^2}{3} $

$\Rightarrow 6 y^2(e)=2 e^2$

Solutions · 1