Question

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

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

Answer 1

Answer: (B)

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