Question

Let $y=y(x)$ be the solution of the differential equation
$\left(3 y^2-5 x^2\right) y d x+2 x\left(x^2-y^2\right) d y=0$ such that $y(1)=1$. Then $\left|(y(2))^3-12 y(2)\right|$ is equal to :

• A. 64
• B. $32 \sqrt{2}$
• C. 32
• D. $16 \sqrt{2}$

Answer 1

Answer: (B)

$\left(3 y^2-5 x^2\right) y \cdot d x+2 x\left(x^2-y^2\right) d y=0$
$ \Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^2-3 y^2\right)}{2 x\left(x^2-y^2\right)} $
Put $y=m x $
$\Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^2\right)}{2\left(1-m^2\right)}$
$x \cdot \frac{d m}{d x}=\frac{\left(5-3 m^2\right) m-2 m\left(1-m^2\right)}{2\left(1-m^2\right)}$
$\Rightarrow \frac{ dx }{ x }=\frac{2\left( m ^2-1\right)}{ m \left( m ^2-3\right)} dm$
$\Rightarrow \frac{ dx }{ x }=\left(\frac{2}{ m }-\frac{\frac{4}{3}}{ m }+\frac{\frac{4 m }{3}}{ m ^2-3}\right) dm$
$ \Rightarrow \int \frac{ dx }{ x }=\int \frac{\left(\frac{2}{3}\right)}{ m }+\int \frac{2}{3}\left(\frac{2 m }{ m ^2-3}\right) dm $
$ \Rightarrow \ln | x |=\frac{2}{3} \ln | m |+\frac{2}{3} \ln \left| m ^2-3\right|+ C $
$ \text { Or, } \ln | x |=\frac{2}{3} \ln \left|\frac{ y }{ x }\right|+\frac{2}{3} \ln \left|\left(\frac{ y }{ x }\right)^2-3\right|+ C $
$ \text { Put }( x =1, y =1): \text { we get } c =-\frac{2}{3} \ln (2)$
$ \Rightarrow \ln | x |=\frac{2}{3} \ln \left|\frac{ y }{ x }\right|+\frac{2}{3} \ln \left|\left(\frac{ y }{ x }\right)^2-3\right|-\frac{2}{3} \ln (2)$
$ \Rightarrow\left(\frac{ y }{ x }\right)\left[\left(\frac{ y }{ x }\right)^2-3\right]=2 .\left( x ^{3 / 2}\right)$
Put $x=2$ to get $y$ ( 2 )
$ \Rightarrow y \left( y ^2-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2} $
$ \Rightarrow y ^3-12 y =32 \sqrt{2}$
$ \Rightarrow\left| y ^3(2)-12 y (2)\right|=32 \sqrt{2}$