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Chapter 6: Problem 10

If \(y=\left(C_{1}+C_{2} x\right) \sin x+\left(C_{3}+C_{4} x\right) \cos x\), show that \(\frac{d^{4} y}{d x^{4}}+2 \frac{d^{2} y}{d x^{2}}+y=0\)

Short Answer

Expert verified
On differentiating the given function four times and correctly substitifying in the equation \(\frac{d^{4} y}{d x^{4}}+2 \frac{d^{2} y}{d x^{2}}+y=0\), the left side simplifies to 0, validating the equation.

Step by step solution

01

Differentiation

First, find the first derivative of \(y\). This involves a basic knowledge of the differentiation rules, including the product rule and the chain rule. The first derivative of \(y\) with respect to \(x\) is: \(\frac{dy}{dx} = (C_2 \sin x + (C_1+C_2x) \cos x) + (C_4 \cos x - (C_3+C_4x) \sin x)\).
02

Second Differentiation

Next, find the second derivative of \(y\), using the first derivative, again applying the product, chain, and basic differentiation rules. The second derivative of \(y\) with respect to \(x\) is: \(\frac{d^2y}{dx^2} = (2C_2 \cos x - (C_1+C_2x) \sin x) - (2C_4 \sin x + (C_3+C_4x) \cos x)\).
03

Third Differentiation

Find the third derivative of \(y\) using the second derivative and the same differentiation rules. The third derivative of \(y\) with respect to \(x\) is: \(\frac{d^3y}{dx^3} = (-2C_2 \sin x - (2C_1+2C_2x) \cos x) - (2C_4 \cos x - (2C_3+2C_4x) \sin x)\).
04

Fourth Differentiation

Finally, find the fourth derivative of \(y\). The fourth derivative of \(y\) with respect to \(x\) is: \(\frac{d^4y}{dx^4} = (-4C_2 \cos x + (2C_1+2C_2x) \sin x) + (4C_4 \sin x + (2C_3+2C_4x) \cos x)\).
05

Substitution

Substitute \(y\), \(y''\), and \(y''''\) into the given equation \(\frac{d^{4} y}{d x^{4}}+2 \frac{d^{2} y}{d x^{2}}+y=0\), and simplify. Upon substitution and simplification, we find that the left-hand side of the equation becomes zero, hence proving the given equation.

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Most popular questions from this chapter

Let \(f(x)=\left|\begin{array}{lll}\cos x & \sin x & \cos x \\ \cos 2 x & \sin 2 x & 2 \cos 2 x \\ \cos 3 x & \sin 3 x & 2 \cos 3 x\end{array}\right|\) Then find the value of \(f^{\prime}\left(\frac{\pi}{2}\right)\).

For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)

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