Question

The solution of the equation $$\frac{{dy}}{{dx}} = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} $$    is :

A. $${\sin ^{ - 1}}y - {\sin ^{ - 1}}x = c$$  
B. $${\sin ^{ - 1}}y\,{\sin ^{ - 1}}x = c$$
C. $${\sin ^{ - 1}}\left( {xy} \right) = 2$$
D. none of these
Answer :   $${\sin ^{ - 1}}y - {\sin ^{ - 1}}x = c$$
Solution :
$$\eqalign{ & \frac{{dy}}{{dx}} = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} \cr & \therefore \,\frac{{dy}}{{\sqrt {1 - {y^2}} }} = \frac{{dx}}{{\sqrt {1 - {x^2}} }} \cr & \Rightarrow \int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr & \Rightarrow {\sin ^{ - 1}}y = {\sin ^{ - 1}}x + c \cr & \therefore \,{\sin ^{ - 1}}y - {\sin ^{ - 1}}x = c \cr} $$

Releted MCQ Question on
Calculus >> Differential Equations

Releted Question 1

A solution of the differential equation $${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$$     is-

A. $$y=2$$
B. $$y=2x$$
C. $$y=2x-4$$
D. $$y = 2{x^2} - 4$$
Releted Question 2

If $${x^2} + {y^2} = 1,$$   then

A. $$yy'' - 2{\left( {y'} \right)^2} + 1 = 0$$
B. $$yy'' + {\left( {y'} \right)^2} + 1 = 0$$
C. $$yy'' + {\left( {y'} \right)^2} - 1 = 0$$
D. $$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$
Releted Question 3

If $$y\left( t \right)$$ is a solution $$\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1$$    and $$y\left( 0 \right) = - 1,$$   then $$y\left( 1 \right)$$ is equal to-

A. $$ - \frac{1}{2}$$
B. $$e + \frac{1}{2}$$
C. $$e - \frac{1}{2}$$
D. $$\frac{1}{2}$$
Releted Question 4

If $$y = y\left( x \right)$$   and $$\frac{{2 + \sin \,x}}{{y + 1}}\left( {\frac{{dy}}{{dx}}} \right) = - \cos \,x,\,y\left( 0 \right) = 1,$$
then $$y\left( {\frac{\pi }{2}} \right)$$   equals-

A. $$\frac{1}{3}$$
B. $$\frac{2}{3}$$
C. $$ - \frac{1}{3}$$
D. $$1$$

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Differential Equations


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