The solution of (d y)/(d x)= 1-x^2-y^2+x^2 y^2 is where c is an arbitrary constant.

Correct answer: C. 2 ^ -1 y=x 1-x^2 + ^ -1 x+c aligned & (d y)/(d x)= 1-x^2-y^2+x^2 y^2 & (d y)/(d x)= (1-x^2)(1-y^2) & d y 1-y^2 = 1-x^2 · d x & =∫ d y 1-y^2 =∫ 1-x^2 · d x [integrating b/s] & = ^ -1 ((y)/(1))=(x)/(2) 1-x^2 +(1)/(2) ^ -1 ((x)/(1))+c & =2 ^ -1 y=x 1-x^2 + ^ -1 x+c aligned

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Differential Equations

NDA 2015 · Mathematics · Differential Equations

The solution of (d y)/(d x)= 1-x^2-y^2+x^2 y^2 is where c is an...

The solution of

\frac{d y}{d x}=\sqrt{1-x^2-y^2+x^2 y^2}

is where c is an arbitrary constant.

Options

A $\sin ^{-1} \mathrm{y}=\sin ^{-1} \mathrm{x}+\mathrm{c}$
B $2 \sin ^{-1} y=\sqrt{1-x^2}+\sin ^{-1} x+c$
C $2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+c$
D $2 \sin ^{-1} y=x \sqrt{1-x^2}+\cos ^{-1} x+c$

Correct answer

C. $2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+c$

Step-by-step solution

\begin{aligned} & \because \frac{d y}{d x}=\sqrt{1-x^2-y^2+x^2 y^2} \\ & \frac{d y}{d x}=\sqrt{\left(1-x^2\right)\left(1-y^2\right)} \\ & \Rightarrow \frac{d y}{\sqrt{1-y^2}}=\sqrt{1-x^2} \cdot d x \\ & =\int \frac{d y}{\sqrt{1-y^2}}=\int \sqrt{1-x^2} \cdot d x \quad \quad \text{ [integrating b/s] } \\ & =\sin ^{-1}\left(\frac{y}{1}\right)=\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}\left(\frac{x}{1}\right)+c \\ & =2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+c\end{aligned}

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