Q120 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics

The solution of the differential equation $\sqrt{1-y^2} d x+x d y-\sin ^{-1} y d y=0$, is

  • Correct answer
    A
    $x=\sin ^1 y-1+c e^{\sin ^{-1} y}$
  • B
    $y=x \sqrt{1-y^2}+\sin ^1 y+c$
  • C
    $x=1+\sin ^1 y+c e^{\sin ^{-1} y}$
  • D
    $y=\sin ^1 y-1+x \sqrt{1-y^2}+c$

Solution

We have, $$ \begin{aligned} & \sqrt{1-y^2} d x+x d y-\sin ^{-1} y d y=0 \\ & \frac{d x}{d y}+\frac{x}{\sqrt{1-y^2}}=\frac{\sin ^{-1} y}{\sqrt{1-y^2}} \\ & \operatorname{IF}=e^{\int \frac{d y}{\sqrt{1-y^2}}}=e^{\sin ^{-1} y} \end{aligned} $$ Solution of given differential equation is $$ \begin{aligned} x e^{\sin ^{-1} y} & =\int e^{\sin ^{-1} y} \frac{\sin ^{-1} y}{\sqrt{1-y^2}} d y \\ x e^{\sin ^{-1} y} & =e^{\sin ^{-1} y}\left(\sin ^{-1} y-1\right)+c \\ x & =\sin ^{-1} y-1+c e^{-\sin ^{-1} y} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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