Question

If a curve satisfying $x{y}_1-4y-x^2\sqrt{y}=0$ passes through $(1,(\log 4{)}^2)$, then the value of $\frac{y\left(2\right)}{(\log 32{)}^2}$ is equal to

• A. 2
• B. 4
• C. 8
• D. 10

Answer 1

The correct option is B 4

$x\frac{dy}{dx}-4y-x^2\sqrt{y}=0\Rightarrow \frac{\frac{dy}{dx}}{2\sqrt{y}}-\frac{2\sqrt{y}}{x}=\frac{x}{2}$
Let $v=\sqrt{y}\Rightarrow \frac{dv}{dx}=\frac{\frac{dy}{dx}}{2\sqrt{y}}$
$\frac{dv}{dx}-\frac{2v}{x}=\frac{x}{2}$ ...(1)
Now let $u=e^{\int -\frac{2}{x}dx}=\frac{1}{x^2}$
Multiplying both sides of (1) by $u$
$\frac{\frac{dv}{dx}}{x^2}-\frac{2v}{x^3}=\frac{1}{2x}\Rightarrow \frac{\frac{dv}{dx}}{x^2}+\frac{d}{dx}\left(\frac{1}{x^2}\right)=\frac{1}{2x}$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d\left(fg\right)}{dx}$
$\frac{d}{dx}\left(\frac{v}{x^2}\right)=\frac{1}{2x}$
Integrating w.r.t x
$\frac{v}{x^2}=\frac{\log x}{2}+c\Rightarrow y=\frac{1}{4}{x}^4{(\log x+2c)}^2$
As it passes through $(1,{\left(\log 4\right)}^2)$
${\left(\log 4\right)}^2=\frac{1}{4}{(\log 1+2c)}^2\Rightarrow c=\log 4$
Now for $x=2$
$y\left(2\right)=\frac{1}{4}{2}^4{(\log 2+2\log 4)}^2=4{\left(\log 32\right)}^2\Rightarrow \frac{y\left(2\right)}{{\left(\log 32\right)}^2}=4$