Question
The particular solution of the differential
equation $${\sin ^{ - 1}}\left( {\frac{{{d^2}y}}{{d{x^2}}} - 1} \right) = x,$$ where $$y = \frac{{dy}}{{dx}} = 0$$ when $$x = 0,$$ is :
A.
$$y = {x^2} + x - \sin \,x$$
B.
$$y = \frac{{{x^2}}}{2} + x - \sin \,x$$
C.
$$y = \frac{{{x^2}}}{2} + \frac{x}{2} - \sin \,x$$
D.
$$2y = {x^2} + x - \sin \,x$$
Answer :
$$y = \frac{{{x^2}}}{2} + x - \sin \,x$$
Solution :
The differential equation is $$\frac{{{d^2}y}}{{d{x^2}}} = 1 + \sin \,x.....\left( {\text{i}} \right)$$
Integrating we get $$\frac{{dy}}{{dx}} = x - \cos \,x + c......\left( {{\text{ii}}} \right)$$
When $$x = 0,\,\frac{{dy}}{{dx}} = 0 \Rightarrow c = 1$$
$$\therefore $$ Equation $$\left( {{\text{ii}}} \right)$$ is $$\frac{{dy}}{{dx}} = x - \cos \,x + 1$$
Integrating again we get $$y = \frac{{{x^2}}}{2} - \sin \,x + x + D......\left( {{\text{iii}}} \right)$$
When $$x = 0,\,y = 0\, \Rightarrow D = 0$$
$$\therefore $$ The particular solution is $$y = \frac{{{x^2}}}{2} + x - \sin \,x$$