The general solution of ((d y/d x))2=1-x2-y2+x2 y2 is

Question

The general solution of ((d y/d x))2=1-x2-y2+x2 y2 is (A) 2 sin-1 y=x √1-y2+c (B) cos-1 y=x cos-1 x+c (C) Sin-1y=(1/2)Sin-1x+c (D) 2 sin-1 y=x √1-x2+sin-1 x+c

Tardigrade Admin · Accepted Answer

Given, ((d y/d x))2=1-x2-y2+x2 y2 ⇒ ((d y/d x))2=(1-x2)-y2(1-x2) ⇒ (d y/d x)=√(1-x2) √(1-y2) ∫ (d y/√(1-y2))=∫ √(1-x2) d x (on integrating) ⇒ sin -1 y=(x/2) √1-x2+(1/2) sin -1 x+(C/2) ⇒ 2 sin -1 y=x √1-x2+ sin -1 x+C

Q. The general solution of $(\frac{d y}{d x})^{2} = 1 - x^{2} - y^{2} + x^{2} y^{2}$ is

$2 s i n^{- 1} y = x 1 - y^{2} + c$

$co s^{- 1} y = x co s^{- 1} x + c$

$S i n^{- 1} y = \frac{1}{2} S i n^{- 1} x + c$

$2 s i n^{- 1} y = x 1 - x^{2} + s i n^{- 1} x + c$

Q. The general solution of (dxdy​)2=1−x2−y2+x2y2 is

2sin−1y=x1−y2​+c

cos−1y=xcos−1x+c

Sin−1y=21​Sin−1x+c

2sin−1y=x1−x2​+sin−1x+c

Given, (dxdy​)2=1−x2−y2+x2y2 ⇒(dxdy​)2=(1−x2)−y2(1−x2) ⇒dxdy​=(1−x2)​(1−y2)​ ∫(1−y2)​dy​=∫(1−x2)​dx (on integrating) ⇒sin−1y=2x​1−x2​+21​sin−1x+2C​ ⇒2sin−1y=x1−x2​+sin−1x+C

Q. The general solution of $(\frac{d y}{d x})^{2} = 1 - x^{2} - y^{2} + x^{2} y^{2}$ is

$2 s i n^{- 1} y = x 1 - y^{2} + c$

$co s^{- 1} y = x co s^{- 1} x + c$

$S i n^{- 1} y = \frac{1}{2} S i n^{- 1} x + c$

$2 s i n^{- 1} y = x 1 - x^{2} + s i n^{- 1} x + c$