Differential Equations Question 307
Question: The particular solution of the differential equation $ {{\sin }^{-1}}( \frac{d^{2}y}{dx^{2}}-1 )=x $ , where $ y=\frac{dy}{dx}=0 $ when $ x=0 $ , is
Options:
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 $
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Answer:
Correct Answer: B
Solution:
[b] The differential equation is $ \frac{d^{2}y}{dx^{2}}=1+\sin x $
……. (i) Integrating we get $ \frac{dy}{dx}=x-\cos x+c $ - (ii) When $ x=0,\frac{dy}{dx}=0\Rightarrow c=1 $
$ \therefore $ Equation (ii) is $ \frac{dy}{dx}=x-\cos x+1 $ Integrating again we get $ y=\frac{x^{2}}{2}-\sin x+x+D $
- (iii) When $ x=0,y=0\Rightarrow D=0 $
$ \therefore $ The particular solution is $ y=\frac{x^{2}}{2}+x-sinx $
BETA
