Question

Let $y=y(x)$ be the solution of the differential equation $\left(x-x^3\right)dy=\left(y+y{x}^2-3x^4\right)dx,x>2$ lf $y(3)=3$, then $y(4)$ is equal to :

• A. 4
• B. 12
• C. 8
• D. 16

Answer 1

Answer: (B)

$\left(x-x^3\right)dy=\left(y+y{x}^2-3x^4\right)dx$
$\Rightarrow xdy-ydx=\left(y{x}^2-3x^4\right)dx+x^{3}dy$
$\Rightarrow \frac{xdy-ydx}{x^2}=(ydx+xdy)-3x^{2}dx$
$\Rightarrow d\left(\frac{y}{x}\right)=d(xy)-d\left(x^3\right)$
Integrate $\Rightarrow \frac{y}{x}=xy-x^3+c$
given $f(3)=3$
$\Rightarrow \frac{3}{3}=3\times 3-3^3+c$
$\Rightarrow c=19$
$\therefore \frac{y}{x}=xy-x^3+19$
at $x=4,\frac{y}{4}=4y-64+19$
$15y=4\times 45$
$\Rightarrow y=12$