The solution of {{{{dy}}}}{{{{dx}}}} = {1 - {{{x}}² - {{{y}}² + {{{x}}²{{{y}}²} is

Correct Answer is : 2sin⁡−1y=x1−x2+sin⁡−1x+c2{ ^{ - 1}}{{y}} = {{x}} {1 - {{{x}}²} + { ^{ - 1}}{{x}} + {{c}}2sin−1y=x1−x2+sin−1x+c

2{ ^{ - 1}}{{y}} = {1 - {{{x}}²} + { ^{ - 1}}{{x}} + {{c}}

2{ ^{ - 1}}{{y}} = {{x}} {1 - {{{x}}²} + { ^{ - 1}}{{x}} + {{c}}

2{ ^{ - 1}}{{y}} = {{x\;}} {1 - {{{x}}²} + { ^{ - 1}}{{x}} + {{c}}

Correct Answer is : 2sin⁡−1y=x1−x2+sin⁡−1x+c2{ ^{ - 1}}{{y}} = {{x}} {1 - {{{x}}²} + { ^{ - 1}}{{x}} + {{c}}2sin−1y=x1−x2+sin−1x+c

Test Index

UPSC NDA II 2015 Mathematics Paper

Question : 37 of 120

Marks: +1, -0

The solution of

\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \sqrt {1 - {{\rm{x}}^2} - {{\rm{y}}^2} + {{\rm{x}}^2}{{\rm{y}}^2}}

$sin^{-1} y = sin^{-1} x + c$
$2{\sin ^{ - 1}}{\rm{y}} = \sqrt {1 - {{\rm{x}}^2}} + {\sin ^{ - 1}}{\rm{x}} + {\rm{c}}$
$2{\sin ^{ - 1}}{\rm{y}} = {\rm{x}}\sqrt {1 - {{\rm{x}}^2}} + {\sin ^{ - 1}}{\rm{x}} + {\rm{c}}$
$2{\sin ^{ - 1}}{\rm{y}} = {\rm{x\;}}\sqrt {1 - {{\rm{x}}^2}} + {\cos ^{ - 1}}{\rm{x}} + {\rm{c}}$

Solution:

Concept:

Use variable separation and standard integration formulas:

\int \frac{dy}{\sqrt{1-y^{2}}} = \sin^{-1}y + C

and

\int \sqrt{1-x^{2}} \, dx = \frac{x}{2}\sqrt{1-x^{2}} + \frac{1}{2}\sin^{-1}x + C

Explanation:

Step 1: Factor the right-hand side.

Given

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

Observe:

1 - x^{2} - y^{2} + x^{2}y^{2} = (1 - x^{2})(1 - y^{2})

Thus

\frac{dy}{dx} = \sqrt{(1 - x^{2})(1 - y^{2})} = \sqrt{1 - x^{2}} \cdot \sqrt{1 - y^{2}}

Step 2: Separate the variables.

Rearrange:

\frac{dy}{\sqrt{1 - y^{2}}} = \sqrt{1 - x^{2}} \, dx

Step 3: Integrate both sides.

\int \frac{dy}{\sqrt{1 - y^{2}}} = \int \sqrt{1 - x^{2}} \, dx

Left side:

\int \frac{dy}{\sqrt{1 - y^{2}}} = \sin^{-1}y

Right side:

\int \sqrt{1 - x^{2}} \, dx = \frac{x}{2}\sqrt{1 - x^{2}} + \frac{1}{2}\sin^{-1}x

Step 4: Combine results with constant of integration.

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

Multiply both sides by 2:

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

Answer:

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

matches option C.

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