Question

Which of the following does not represent the orthogonal trajectory of the system of curves $${\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{a}{x}$$

A. $$9a{\left( {y + c} \right)^2} = 4{x^3}$$
B. $$y + c = \frac{{ - 2}}{{3\sqrt a }}{x^{\frac{3}{2}}}$$
C. $$y + c = \frac{2}{{3\sqrt a }}{x^{\frac{3}{2}}}$$
D. All are orthogonal trajectories  
Answer :   All are orthogonal trajectories
Solution :
The family of curves which are orthogonal (i.e. intersect at right angles) to a given system of curves is obtained by substituting $$ - \frac{{dx}}{{dy}}$$  for $$\frac{{dy}}{{dx}}$$  in the differential equation of the given system.
The given differential equation is $${\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{a}{x}$$
Replacing $$\frac{{dy}}{{dx}}$$  by $$ - \frac{{dx}}{{dy}},$$  we get
$${\left( {\frac{{dx}}{{dy}}} \right)^2} = \frac{a}{x}\, \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{x}{a}\, \Rightarrow \frac{{dy}}{{dx}} = \pm \sqrt {\frac{x}{a}} $$
Integrating we get
$$\eqalign{ & y + c = \pm \frac{2}{{3\sqrt a }}{x^{\frac{3}{2}}}......\left( {\text{i}} \right) \cr & \Rightarrow {\left( {y + c} \right)^2} = \frac{4}{{9a}}{x^3} \cr & \Rightarrow 9a{\left( {y + c} \right)^2} = 4{x^3}......\left( {{\text{ii}}} \right) \cr} $$
From $$\left( {{\text{i}}} \right)$$ and $$\left( {{\text{ii}}} \right)$$ all of the first three given options represent required equations.

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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