Question

Given, $y=1+\cos x\&y=1+\sin x$ are solutions of the differential equation, $\frac{d^{2}y}{d{x}^2}+y=1$, then its solution will be also :

• A. $y=2(1+\cos x)$
• B. $y=2+\cos x+\sin x$
• C. $y=\cos x-\sin x$
• D. $y=1+\cos x+\sin x$

Answer 1

Answer: (D)

Since, $y_1=1+\cos x$ and $y_2=1+\sin x$ are the solution of the $DE\frac{d^{2}y}{d{x}^2}+y=1$
hence $\frac{d^2{y}_1}{d{x}^2}+y_1=1$ and $\frac{d^2{y}_2}{d{x}^2}+y_2=1$; on adding, we get
$\frac{d^2}{d{x}^2}\left(y_1+y_2\right)+\left(y_1+y_2\right)=2$
or $\frac{d^2}{d{x}^2}\left(y_1+y_2-1\right)+\left(y_1+y_2-1\right)=1$ ....(1)
conparing this with $\frac{d^{2}y}{d{x}^2}+y=1$, we have $y_1+y_2-1=0$ also satisfies the $DE$
$\therefore 1+\cos x+\sin x=0\Rightarrow$ (D)