KCET · Maths · Differential Equations
The general solution of \(\left(\frac{d y}{d x}\right)^{2}=1-x^{2}-y^{2}+x^{2} y^{2}\) is
- A \(2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+C\)
- B \(\cos ^{-1} y=x \cos ^{-1} x+C\)
- C \(\sin ^{-1} y=\frac{1}{2} \sin ^{-1} x+C\)
- D \(2 \sin ^{-1} y=x \sqrt{1-y^{2}}+C\)
Answer & Solution
Correct Answer
(A) \(2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+C\)
Step-by-step Solution
Detailed explanation
Given, \(\quad\left(\frac{d y}{d x}\right)^{2}=1-x^{2}-y^{2}+x^{2} y^{2}\)
\(\Rightarrow \quad\left(\frac{d y}{d x}\right)^{2}=\left(1-x^{2}\right)-y^{2}\left(1-x^{2}\right)\)
\(\Rightarrow \quad \frac{d y}{d x}=\sqrt{\left(1-x^{2}\right)} \sqrt{\left(1-y^{2}\right)}\)
\(\Rightarrow \frac{d y}{\sqrt{\left(1-y^{2}\right)}}=\int \sqrt{\left(1-x^{2}\right)} d x\) (on integrating)
\(\Rightarrow \quad \sin ^{-1} y=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x+\frac{C}{2}\)
\(\Rightarrow \quad 2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+C\)
\(\Rightarrow \quad\left(\frac{d y}{d x}\right)^{2}=\left(1-x^{2}\right)-y^{2}\left(1-x^{2}\right)\)
\(\Rightarrow \quad \frac{d y}{d x}=\sqrt{\left(1-x^{2}\right)} \sqrt{\left(1-y^{2}\right)}\)
\(\Rightarrow \frac{d y}{\sqrt{\left(1-y^{2}\right)}}=\int \sqrt{\left(1-x^{2}\right)} d x\) (on integrating)
\(\Rightarrow \quad \sin ^{-1} y=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x+\frac{C}{2}\)
\(\Rightarrow \quad 2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+C\)
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