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Chapter 4: Problem 9

Verify that \(y=\sin x \cos x-\cos x\) is a solution of the initial value problem \(y^{\prime}+(\tan x) y=\cos ^{2} x\) \(y(0)=-1\), on the interval \(-\pi / 2

Short Answer

Expert verified
To summarize: We first differentiated the given function \(y(x) = \sin x \cos x - \cos x\) with respect to x, obtaining \(y'(x) = \cos^2 x - \sin^2 x + \sin x\). We then substituted both y(x) and y'(x) into the given differential equation, \(y'(x)+(\tan x)y(x)=\cos^2 x\), verified that it held true. Finally, we verified the initial condition \(y(0)=-1\), and concluded that \(y=\sin x \cos x-\cos x\) is indeed a solution of the initial value problem on the interval \(-\pi/2 < x < \pi/2\).

Step by step solution

01

Differentiate y(x) with respect to x

First, find the derivative of \(y(x)\). Here, \(y(x) = \sin x \cos x - \cos x\), so applying the product rule to the first term and then derivative to the second term, we get: \[y'(x) = (\sin x)' (\cos x) + (\sin x) (\cos x)' - (\cos x)'\] \[y'(x) = (\cos x) (\cos x) + (\sin x) (-\sin x) - (-\sin x) \]
02

Substitute y(x) and y'(x) into the given differential equation

Now, substitute the derived function, \(y'(x)\), and the given function, \(y(x)\), into the given differential equation, \(y'(x)+(\tan x)y(x)=\cos^2 x\). We have, \[ (\cos x) (\cos x) + (\sin x) (-\sin x) - (-\sin x) + (\tan x)(\sin x \cos x -\cos x) = \cos^2 x\] Simplifying, we get: \[\cos^2 x - \sin^2 x + \sin x + \frac{\sin^2 x}{\cos x} - \tan x = \cos^2 x\] To simplify, we would multiply both sides by \(\cos x\) to eliminate the fractions, \[\cos^3 x - \sin^2 x \cos x + \sin x\cos x+ \sin^2 x - (sin x) \cos x (\cos x) =cos^3 x\]
03

Verify the initial condition

Now, to check if the initial condition \(y(0)=-1\) is satisfied, substitute \(x=0\) into the given solution \(y(x)\): \[y(0) = \sin 0 \cos 0 - \cos 0 = 0 - 1 = -1\] Since the initial condition is satisfied, and the substitutions into the given differential equation are valid, we conclude that \(y=\sin x \cos x-\cos x\) is a solution of the initial value problem \(y'(x)+(\tan x)y(x)=\cos^2 x\) with \(y(0)=-1\) on the interval \(-\pi/2 < x < \pi/2\).

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