the order and degree of the differential equation of the

The order and degree of the differential equation of the family of circles touching the $$x$$-axis at the origin, are respectively :

A. $$1,\,1$$

B. $$1,\,2$$

C. $$2,\,1$$

D. $$2,\,2$$

Answer : $$1,\,1$$

Solution :
The equation of the family is
$$\eqalign{ & {\left( {x - c} \right)^2} + {y^2} = {c^2}{\text{ or }}{x^2} + {y^2} - 2cx = 0 \cr & {\text{or }}\frac{{{x^2} + {y^2}}}{x} = 2c\,\,\,\,\,\,\,\,\, \Rightarrow \frac{{\left( {2x + 2y\frac{{dy}}{{dx}}} \right)x - \left( {{x^2} + {y^2}} \right).1}}{{{x^2}}} = 0 \cr} $$
So, degree $$=1$$ order $$=1$$

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

The order and degree of the differential equation of the family of circles touching the $$x$$-axis at the origin, are respectively :

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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The order and degree of the differential equation of the family of circles touching the $$x$$-axis at the origin, are respectively :

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 $$y = y\left( x \right)$$ 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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