Question

Let $y=y(x)$ be the solution of the differential equation $x\left(1-x^2\right)\frac{dy}{dx}+\left(3x^{2}y-y-4x^3\right)=0,x>1$ with $y(2)=-2$. Then $y(3)$ is equal to

• A. $-18$
• B. $-12$
• C. $-6$
• D. $-3$

Answer 1

Answer: (A)

$x\left(1-x^2\right)\frac{dy}{dx}+\left(3x^{2}y-y-4x^3\right)=0$
$x\left(1-x^2\right)\frac{dy}{dx}+\left(3x^2-1\right)y=4x^3$
$\frac{dy}{dx}+\frac{\left(3x^2-1\right)}{\left(x-x^3\right)}y=\frac{4x^3}{\left(x-x^3\right)}$
$\frac{dy}{dx}+Py=Q$
$IF=e^{\int Pdx}=e^{\int \frac{3x^2-1}{x-x^3}dx}$
$x-x^3=t\Rightarrow IF=e^{\int \frac{-dt}{t}}$
$=e^{-\text{lnt}}=\frac{1}{t}$
$\therefore IF=\frac{1}{x-x^3}$
$y\times IF=\int Q\times IFdx$
$y\left(\frac{1}{x-x^3}\right)=\int \frac{4x^3}{x-x^3}\times \frac{1}{\left(x-x^3\right)}dx$
$=\int \frac{4x^3}{{\left(x-x^3\right)}^2}dx$
$=\int \frac{4x}{{\left(1-x^2\right)}^2}dx],1-x^2=K$
$=2\int \frac{-dK}{K^2}],-2xdx=dK$
$=-2\left(-\frac{1}{K}\right)+c$
$\frac{y}{x-x^3}=\frac{2}{K}+c$
$\frac{y}{x-x^3}=\frac{2}{1-x^2}+c$
At $x=2,y=-2$
$\frac{-2}{2-8}=\frac{2}{1-4}+c$
$\frac{1}{3}=\frac{-2}{3}+c$
$\therefore C=1$
$\frac{y}{x-x^3}=\frac{2}{1-x^2}+1$
Put $x=3$
$\frac{y}{3-27}=\frac{2}{1-9}+1$
$\frac{y}{-24}=-\frac{1}{4}+1$
$\frac{y}{-24}=\frac{3}{4}$
$y=\frac{3}{4}(-24)=-18$