Question

Solution of the differential equation $$x = 1 + xy\frac{{dy}}{{dx}} + \frac{{{x^2}{y^2}}}{{2!}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{{{x^3}{y^3}}}{{3!}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + ......$$

A. $$y = \ln \left( x \right) + c$$
B. $$y = {\left( {\ln \,x} \right)^2} + c$$
C. $$y = \pm \ln \left( x \right) + c$$  
D. $$xy = {x^y} + c$$
Answer :   $$y = \pm \ln \left( x \right) + c$$
Solution :
The given equation is reduced to
$$\eqalign{ & x = {e^{xy\left( {\frac{{dy}}{{dx}}} \right)}} \cr & \Rightarrow \ell n\,x = xy\frac{{dy}}{{dx}} \cr & \Rightarrow \int {y\,dy} = \int {\frac{1}{x}} \ell n\,x\,dx \cr & \Rightarrow \frac{{{y^2}}}{2} = \frac{{{{\left( {\ell n\,x} \right)}^2}}}{2} + c \cr & \Rightarrow y = \pm \sqrt {{{\left( {\ell n\,x} \right)}^2}} + c \cr & \Rightarrow y = \pm \ell n\,x + c \cr} $$

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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