Question

If $\frac{dy}{dx}=\left(x^3-2x{\left(sin\right)}^{-1}y\right)\sqrt{1-y^2}$ then its general solution will be

• A. $2{\left(sin\right)}^{-1}y=\left(x^2-1\right)+C{e}^{-x^2}$
• B. $2{\left(cos\right)}^{-1}y=\left(x^4+1\right)+C$
• C. $e^{x^2}{\left(sin\right)}^{-1}y=\left(x^2-x\right)+C$
• D. $2{\left(cos\right)}^{-1}y=\left(x^2-1\right)e^{-x^2}+C$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=x^3-2xsi{n}^{-1}y$
$\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}+2xsi{n}^{-1}y=x^3$
let $si{n}^{-1}y=t$
$\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=\frac{dt}{dx}$

I.F. $=e^{\int 2xdx}=e^{x^2}$
$\therefore$ solution is given by
$t\left(e^{x^2}\right)=\int x^3\cdot e^{x^2}dx+c$
let $x^2=t$
$si{n}^{-1}y=e^{x^2}=\int t{e}^t\cdot \frac{dt}{2}+c2ndx=dt$
$\Rightarrow 2{\left(sin\right)}^{-1}y{e}^{x^2}=\left(t\cdot e^t-\int e^{{}^{\prime }}dt\right)+2C$
$\Rightarrow 2{\left(sin\right)}^{-1}y{e}^{x^2}=\left(x^2-1\right)e^{x^2}+K$
$\Rightarrow 2{\left(sin\right)}^{-1}y=\left(x^2-1\right)+K{e}^{-x^2}$