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) \[2y={{x}^{2}}+x-\sin x\]
Correct Answer: B
Solution :
[b] The differential equation is \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=1+\sin x\] ?. (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 \[y=\frac{{{x}^{2}}}{2}+x-\operatorname{sinx}\]You need to login to perform this action.
You will be redirected in
3 sec