Question

If $y=a\cos x+\left( b+2x \right)\sin x$, then show that ${{y}^{''}}+y=4\cos x$.

Answer 1

Hint: The differentiation of the given function $y=a\cos x+\left( b+2x \right)\sin x$ will be defined as ${{f}^{'}}\left( x \right)$ and ${{f}^{''}}\left( x \right)$ respectively where $y=f\left( x \right)=a\cos x+\left( b+2x \right)\sin x$. We differentiate the given function $y=a\cos x+\left( b+2x \right)\sin x$ with $x$. The differentiated forms are $\dfrac{d}{dx}\left[ \sin \left( x \right) \right]=\cos x$ and $\dfrac{d}{dx}\left[ \cos x \right]=-\sin x$. We use those forms and keep the constants as it is.

Complete step by step answer:
In case of finding any derivative, we have to differentiate the given function $y=a\cos x+\left( b+2x \right)\sin x$ with $x$. Let’s assume that $y=f\left( x \right)=a\cos x+\left( b+2x \right)\sin x$. The given function is a function of $x$.
The first derivative is $\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ f\left( x \right) \right]$. It’s also defined as ${{f}^{'}}\left( x \right)$.
The differentiated forms are $\dfrac{d}{dx}\left[ \sin \left( x \right) \right]=\cos x$ and $\dfrac{d}{dx}\left[ \cos x \right]=-\sin x$. The constant remains as it is. We also have $\dfrac{d}{dx}\left[ {{x}^{n}} \right]=n{{x}^{n-1}}$.
There is a function $af\left( x \right)$ where $a$ is a constant. If we are going to differentiate the function the formula remains as $\dfrac{d}{dx}\left[ af\left( x \right) \right]=a\dfrac{d}{dx}\left[ f\left( x \right) \right]$.
Therefore, the first derivative of $y=a\cos x+\left( b+2x \right)\sin x$ is
$\begin{align}
  & {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ a\cos x+\left( b+2x \right)\sin x \right] \\
 & =-a\sin x+\left( b+2x \right)\cos x+2\sin x \\
 & =\left( b+2x \right)\cos x+\left( 2-a \right)\sin x \\
\end{align}$.
Now we need to find the second derivative.
The second derivative is $\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]$. It’s also defined as ${{f}^{''}}\left( x \right)$.
The differentiation of constants has been discussed before.
Therefore, the second derivative of ${{f}^{'}}\left( x \right)=\left( b+2x \right)\cos x+\left( 2-a \right)\sin x$ is
\[\begin{align}
  & {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left[ \left( b+2x \right)\cos x+\left( 2-a \right)\sin x \right] \\
 & =-\left( b+2x \right)\sin x+2\cos x+\left( 2-a \right)\cos x \\
 & =\left( 4-a \right)\cos x-\left( b+2x \right)\sin x \\
\end{align}\]
So, we get ${{y}^{''}}={{f}^{''}}\left( x \right)=\left( 4-a \right)\cos x-\left( b+2x \right)\sin x$ and $y=a\cos x+\left( b+2x \right)\sin x$.
Adding them we get
$\begin{align}
  & {{y}^{''}}+y \\
 & =\left( 4-a \right)\cos x-\left( b+2x \right)\sin x+a\cos x+\left( b+2x \right)\sin x \\
 & =\left( 4-a+a \right)\cos x \\
 & =4\cos x \\
\end{align}$
Thus proved, ${{y}^{''}}+y=4\cos x$.

Note: The second derivative can also be expressed as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$. The differentiation $\dfrac{d}{dx}\left[ \left( b+2x \right)\sin x \right]=\left( b+2x \right)\cos x+2\sin x$ has been done following the rule of chain rule where \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ goh\left( x \right) \right]=\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}={{g}^{'}}\left[ h\left( x \right) \right]{{h}^{'}}\left( x \right)\] for $f\left( x \right)=goh\left( x \right)$.