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

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

Explanation

Solution

The differentiation of the given function y=acos⁡x+(b+2x)sin⁡xy=a\cos x+\left( b+2x \right)\sin x will be defined as f′(x){{f}^{'}}\left( x \right) and f′′(x){{f}^{''}}\left( x \right) respectively where y=f(x)=acos⁡x+(b+2x)sin⁡xy=f\left( x \right)=a\cos x+\left( b+2x \right)\sin x. We differentiate the given function y=acos⁡x+(b+2x)sin⁡xy=a\cos x+\left( b+2x \right)\sin x with xx. The differentiated forms are ddx[sin⁡(x)]=cos⁡x\dfrac{d}{dx}\left[ \sin \left( x \right) \right]=\cos x and ddx[cos⁡x]=−sin⁡x\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=acos⁡x+(b+2x)sin⁡xy=a\cos x+\left( b+2x \right)\sin x with xx. Let’s assume that y=f(x)=acos⁡x+(b+2x)sin⁡xy=f\left( x \right)=a\cos x+\left( b+2x \right)\sin x. The given function is a function of xx.
The first derivative is dydx=ddx[f(x)]\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ f\left( x \right) \right]. It’s also defined as f′(x){{f}^{'}}\left( x \right).
The differentiated forms are ddx[sin⁡(x)]=cos⁡x\dfrac{d}{dx}\left[ \sin \left( x \right) \right]=\cos x and ddx[cos⁡x]=−sin⁡x\dfrac{d}{dx}\left[ \cos x \right]=-\sin x. The constant remains as it is. We also have ddx[xn]=nxn−1\dfrac{d}{dx}\left[ {{x}^{n}} \right]=n{{x}^{n-1}}.
There is a function af(x)af\left( x \right) where aa is a constant. If we are going to differentiate the function the formula remains as ddx[af(x)]=addx[f(x)]\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=acos⁡x+(b+2x)sin⁡xy=a\cos x+\left( b+2x \right)\sin x is
f′(x)=ddx[acos⁡x+(b+2x)sin⁡x] =−asin⁡x+(b+2x)cos⁡x+2sin⁡x =(b+2x)cos⁡x+(2−a)sin⁡x \begin{aligned} & {{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{aligned}.
Now we need to find the second derivative.
The second derivative is ddx[f′(x)]\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]. It’s also defined as f′′(x){{f}^{''}}\left( x \right).
The differentiation of constants has been discussed before.
Therefore, the second derivative of f′(x)=(b+2x)cos⁡x+(2−a)sin⁡x{{f}^{'}}\left( x \right)=\left( b+2x \right)\cos x+\left( 2-a \right)\sin x is

& {{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{aligned}$$ So, we get ${{y}^{''}}={{f}^{''}}\left( x \right)=\left( 4-a \right)\cos x-\left( b+2x \right)\sin x$ and $y=a\cos x+\left( b+2x \right)\sin x$. Adding them we get $\begin{aligned} & {{y}^{''}}+y \\\ & =\left( 4-a \right)\cos x-\left( b+2x \right)\sin x+a\cos x+\left( b+2x \right)\sin x \\\ & =\left( 4-a+a \right)\cos x \\\ & =4\cos x \\\ \end{aligned}$ Thus proved, ${{y}^{''}}+y=4\cos x$. **Note:** The second derivative can also be expressed as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$. The differentiation $\dfrac{d}{dx}\left[ \left( b+2x \right)\sin x \right]=\left( b+2x \right)\cos x+2\sin x$ has been done following the rule of chain rule where $$\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 $f\left( x \right)=goh\left( x \right)$.