Question

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

• A. $2\left(sin\right)^{- 1}y=\left(x^{2} - 1\right)+Ce^{- x^{2}}$
• 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)

$\frac{1}{\sqrt{1 - y^{2}}}\frac{d y}{d x}=x^{3}-2xsin^{- 1}y$
$\frac{1}{\sqrt{1 - y^{2}}}\frac{d y}{d x}+2xsin^{- 1}y=x^{3}$
let $sin^{- 1}y=t$
$\frac{1}{\sqrt{1 - y^{2}}}\frac{d y}{d x}=\frac{d t}{d x}$

I.F. $=e^{\displaystyle \int 2 x d x}=e^{x^{2}}$
$\therefore $ solution is given by
$t\left(e^{x^{2}}\right)=\displaystyle \int x^{3}\cdot e^{x^{2}}dx+c$
let $x^{2}=t$
$sin^{- 1}y=e^{x^{2}}=\displaystyle \int te^{t}\cdot \frac{d t}{2}+c2ndx=dt$
$\Rightarrow 2\left(sin\right)^{- 1}ye^{x^{2}}=\left(t \cdot e^{t} - \displaystyle \int e^{'} d t\right)+2C$
$\Rightarrow 2\left(sin\right)^{- 1}ye^{x^{2}}=\left(x^{2} - 1\right)e^{x^{2}}+K$
$\Rightarrow 2\left(sin\right)^{- 1}y=\left(x^{2} - 1\right)+Ke^{- x^{2}}$