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find the Wronskian of the given pair of functions. $$ \cos ^{2} \theta, \quad 1+\cos 2 \theta $$

Short Answer

Expert verified
The Wronskian of the functions \(\cos ^{2} \theta\) and \(1+\cos 2 \theta \) is given by: $$ W(\cos^2\theta, 1+\cos2\theta) = - 2\sin2\theta\cdot\cos\theta(\cos\theta - \sin\theta) $$

Step by step solution

01

Compute the derivatives of the given functions

Find the first derivatives of the functions \(\cos ^{2} \theta\) and \(1+\cos 2 \theta \). Use the chain rule for the first function and the double angle formula for the second function. $$ \frac{d}{d\theta}(\cos^{2}\theta) = 2\cos\theta \cdot (-\sin\theta) = -2\cos\theta\sin\theta \\ \frac{d}{d\theta}(1+\cos 2\theta) = -2\sin(2\theta) $$
02

Arrange the functions and their derivatives in a determinant

Create a determinant using the given functions and their derivatives. Arrange the functions at the top row and their derivatives at the bottom row. $$ W(\cos^2\theta, 1+\cos2\theta) = \begin{vmatrix} \cos^2\theta & 1+\cos2\theta \\ -2\cos\theta\sin\theta & -2\sin2\theta \end{vmatrix} $$
03

Compute the determinant

To find the Wronskian, compute the determinant of the matrix obtained in Step 2. $$ W(\cos^2\theta, 1+\cos2\theta) = (\cos^2\theta)(-2\sin2\theta) - (1+\cos2\theta)(-2\cos\theta\sin\theta) $$ Now simplify the resulting expression and apply the trigonometric identity \(2\sin\theta\cos\theta = \sin(2\theta)\) $$ W(\cos^2\theta, 1+\cos2\theta)=(\cos^2\theta)(-2\sin2\theta) + 2\cos\theta\sin\theta(1+\cos2\theta) \\ =- 2\sin2\theta(\cos^2\theta - \cos\theta\sin\theta) \\ =- 2\sin2\theta\cdot\cos\theta(\cos\theta - \sin\theta) $$
04

Final Answer

The Wronskian of the given pair of functions is: $$ W(\cos^2\theta, 1+\cos2\theta) = - 2\sin2\theta\cdot\cos\theta(\cos\theta - \sin\theta) $$

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Most popular questions from this chapter

Find the general solution of the given differential equation. $$ y^{\prime \prime}+5 y^{\prime}=0 $$

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