Differential Equations Question 2
Question 2 - 24 January - Shift 2
Let $y=y(x)$ be the solution of the differential equation $(x^{2}-3 y^{2}) d x+3 x y d y=0, y(1)=1$. Then $6 y^{2}(e)$ is equal to
(1) $3 e^{2}$
(2) $e^{2}$
(3) $2e^{2}$
(4) $\frac{3 e^{2}}{2}$
Show Answer
Answer: (3)
Solution:
Formula: Homogeneous equations
$ (x^{2}-3 y^{2}) d x+3 x y d y=0 $
$\frac{d y}{d x}=\frac{3 y^{2}-x^{2}}{3 x y} \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\frac{1}{3} \frac{x}{y}$
Put $y=v x$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore v+x \frac{d v}{d x}=v-\frac{1}{3} \frac{1}{v}$
$\Rightarrow vdv=\frac{-1}{3 x}dx$
Integrating both side
$\frac{v^{2}}{2}=\frac{-1}{3} \ln x+c$
$\Rightarrow \frac{y^{2}}{2 x^{2}}=\frac{-1}{3} \ln x+c$
$y(1)=1$
$\Rightarrow \frac{1}{2}=c$
$\Rightarrow \frac{y^{2}}{2 x^{2}}=\frac{-1}{3} \ln x+\frac{1}{2}$
$ \begin{aligned} & \text{ } \\ & y^{2}(e)=-\frac{2}{3} e^{2}+e^{2}=\frac{e^{2}}{3} \end{aligned} $
