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Q.
Given, $y =1+\cos x \& y =1+\sin x$ are solutions of the differential equation, $\frac{ d ^2 y }{ dx ^2}+ y =1$, then its solution will be also :
Differential Equations
Solution:
Since, $y _1=1+\cos x$ and $y _2=1+\sin x$ are the solution of the $DE \frac{ d ^2 y }{ dx ^2}+ y =1$
hence $\frac{ d ^2 y _1}{ dx ^2}+ y _1=1$ and $\frac{ d ^2 y _2}{ dx ^2}+ y _2=1$; on adding, we get
$\frac{ d ^2}{ dx ^2}\left( y _1+ y _2\right)+\left( y _1+ y _2\right)=2 $
or $ \frac{ d ^2}{ dx ^2}\left( y _1+ y _2-1\right)+\left( y _1+ y _2-1\right)=1$ ....(1)
conparing this with $\frac{ d ^2 y }{ dx ^2}+ y =1$, we have $y _1+ y _2-1=0$ also satisfies the $DE$
$\therefore 1+\cos x+\sin x=0 \Rightarrow$ (D)