Problem 3: Solve the initial value problem given by11y" + y =+1+ cos(x)1+ sec(x)'y(0) = 1, y'(0) = 0

The given differential equation is y''+y=11+cosx+11+secx. The above differential equation can be…

Answered: Problem 3: Solve the initial value…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Problem 3: Solve the initial value problem given by
1
1
y" + y =
+
1+ cos(x)
1+ sec(x)'
y(0) = 1, y'(0) = 0

Expert Solution

Step 1

The given differential equation is $y'' + y = \frac{1}{1 + \cos (x)} + \frac{1}{1 + s e c (x)}$ .

The above differential equation can be rewritten as follows.

$y'' + y = \frac{1}{1 + \cos (x)} + \frac{1}{1 + s e c (x)} \Rightarrow y'' + y = \frac{1}{1 + \cos (x)} + \frac{1}{1 + (\frac{1}{\cos (x)})} \Rightarrow y'' + y = \frac{1}{1 + \cos (x)} + \frac{\cos (x)}{\cos (x) + 1} \Rightarrow y'' + y = \frac{1 + \cos (x)}{1 + \cos (x)} \Rightarrow y'' + y = 1$

The above differential equation is a non-homogeneous differential equation with non-homogeneous part 1.

The corresponding homogeneous differential equation is $y'' + y = 0$ .

Step 2

We know that the characteristic equation of a second order differential equation $a y'' + b y' + c y = 0$ is given by $a m^{2} + b m + c = 0$ .

Hence, the characteristic equation of $y'' + y = 0$ is $m^{2} + 1 = 0$ .

Solve the equation $m^{2} + 1 = 0$ as follows.

$m^{2} + 1 = 0 \Rightarrow m^{2} = - 1 \Rightarrow m = \pm i$

We know that if the solutions of the characteristic equation is of the form $a \pm b i$ , then the general solution of the second order homogeneous differential equation is given by $y = e^{a x} (c_{1} \cos (b x) + c_{2} \sin (b x))$ .

Since $m = \pm i$ , we have $a = 0, b = 1$ .

Hence, the general solution of $y'' + y = 0$ is $y = e^{a x} (c_{1} \cos (b x) + c_{2} \sin (b x))$ .

That is, the homogeneous solution of $y'' + y = 1$ is $y_{h} = c_{1} \cos (x) + c_{2} \sin (x)$ .

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