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If $y=a \sin x+(5+2 x) \cos x$, then $y^{\prime \prime}+y=$
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Verified Answer
The correct answer is:
$-4 \sin x$
$y=a \sin x+(5+2 x) \cos x$
Differential w.r.t ' $x$ '
$$
\begin{aligned}
& y^{\prime}=a \cos x+(5+2 x)(-\sin x)+(0+2) \cos x \\
& y^{\prime}=a \cos x-(5+2 x) \sin x+2 \cos x
\end{aligned}
$$
Again differential w.r.t ' $x$ '
$$
\begin{aligned}
& y^{\prime \prime}=-a \sin x-[(5+2 x) \cos x \\
& \quad+(0+2) \sin x]+2(-\sin x) \\
& \Rightarrow y^{\prime \prime}=-a \sin x-(5+2 x) \cos x-2 \sin x-2 \sin x \\
& \Rightarrow y^{\prime \prime}=-(a \sin x+(5+2 x) \cos x)-4 \sin x \\
& \Rightarrow y^{\prime \prime}=-y-4 \sin x \\
& \Rightarrow y^{\prime \prime}+y=-4 \sin x
\end{aligned}
$$
Differential w.r.t ' $x$ '
$$
\begin{aligned}
& y^{\prime}=a \cos x+(5+2 x)(-\sin x)+(0+2) \cos x \\
& y^{\prime}=a \cos x-(5+2 x) \sin x+2 \cos x
\end{aligned}
$$
Again differential w.r.t ' $x$ '
$$
\begin{aligned}
& y^{\prime \prime}=-a \sin x-[(5+2 x) \cos x \\
& \quad+(0+2) \sin x]+2(-\sin x) \\
& \Rightarrow y^{\prime \prime}=-a \sin x-(5+2 x) \cos x-2 \sin x-2 \sin x \\
& \Rightarrow y^{\prime \prime}=-(a \sin x+(5+2 x) \cos x)-4 \sin x \\
& \Rightarrow y^{\prime \prime}=-y-4 \sin x \\
& \Rightarrow y^{\prime \prime}+y=-4 \sin x
\end{aligned}
$$
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