if x 2 y 2 1 then 747001371

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. $$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$

Answer : $$yy'' + {\left( {y'} \right)^2} + 1 = 0$$

Solution :
Given $${x^2} + {y^2} = 1.$$
Differentiating w.r.t. $$x,$$ we get $$2x + 2yy' = 0\,\,{\text{or }}x + yy' = 0$$
Again differentiating w.r.t. $$x,$$ we get $$1 + y'y' + yy'' = 0\,\,{\text{or 1}} + {\left( {y'} \right)^2} + yy'' = 0$$

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. $$y = 2{x^2} - 4$$

View Answer

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. $$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$

View Answer

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. $$\frac{1}{2}$$

View Answer

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. $$1$$

View Answer

If $${x^2} + {y^2} = 1,$$ then

Releted MCQ Question on
Calculus >> Differential Equations

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

If $${x^2} + {y^2} = 1,$$ then

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-

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-

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If $${x^2} + {y^2} = 1,$$ then

Releted MCQ Question on Calculus >> Differential Equations

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

If $${x^2} + {y^2} = 1,$$ then

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-

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-

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Releted MCQ Question on
Calculus >> Differential Equations

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-

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Differential Equations