Use Laplace transforms to solve the following initial value problem.x"-25x 9t; x(0) = x'(0) = 0

Answered: Use Laplace transforms to solve the…

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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Question

Use Laplace transforms to solve the following initial value problem.
x"-25x 9t; x(0) = x'(0) = 0

Expert Solution

Step 1

Given:

Given IVP is $x'' - 25 x = 9 t, x (0) = x' (0) = 0$

To find solution of IVP we use Laplace transform

Step 2

Solution:

Consider,

$x'' - 25 x = 9 t$

Apply Laplace transform on both side we get

$L \{x'' - 25 x\} = L \{9 t\} \Rightarrow L \{x''\} - 25 L \{x\} = 9 L \{t\} \Rightarrow (s^{2} X (s) - s x (0) - x' (0)) - 25 X (s) = \frac{9}{s^{2}}$

Given initial conditions are $x (0) = x' (0) = 0$ therefore we get

$(s^{2} X (s) - s (0) - 0) - 25 X (s) = \frac{9}{s^{2}} \Rightarrow (s^{2} - 25) X (s) = \frac{9}{s^{2}} \Rightarrow X (s) = \frac{9}{s^{2} (s^{2} - 25)}$

Now we use inverse Laplace transform to find $X (t)$

$X (t) = L^{- 1} \{\frac{9}{s^{2} (s^{2} - 25)}\} \Rightarrow X (t) = L^{- 1} \{\frac{9}{s^{2} (s + 5) (s - 5)}\} . . . . . . . . . . . . . (1)$

Now we take partial fraction

$\frac{9}{s^{2} (s + 5) (s - 5)} = \frac{A}{s^{2}} + \frac{B}{s + 5} + \frac{C}{s - 5} \Rightarrow A (s + 5) (s - 5) + B s^{2} (s - 5) + C s^{2} (s + 5) = 9 \Rightarrow A (s^{2} - 25) + B (s^{3} - 5 s^{2}) + C (s^{3} + 5 s^{2}) = 9 \Rightarrow (B + C) s^{3} + (A - 5 B + 5 C) s^{2} - 25 A = 9$

comparing both side we get system of equations,

$\{\begin{matrix} - 25 A = 9 \Rightarrow A = - \frac{9}{25} \\ A - 5 B + 5 C = 0 \Rightarrow - 5 B + 5 C = \frac{9}{25} . . . . . . . (2) \\ B + C = 0 . . . . . . (3) \end{matrix}$

$- 5 B + 5 C = \frac{9}{25} + 5 B + 5 C = 0 10 C = \frac{9}{25}$

Therefore $C = \frac{9}{250}$ substituting value of C in equation 3 we get

$B + C = 0 \Rightarrow B = - C \Rightarrow B = - \frac{9}{250}$

Therefore Partial fraction of $\frac{9}{s^{2} (s + 5) (s - 5)} = - \frac{9}{25 s^{2}} - \frac{9}{250 (s + 5)} + \frac{9}{250 (s - 5)}$

Hence 1 becomes,

$\begin{array}{rcl} X (t) & = & L^{- 1} \{\frac{9}{s^{2} (s + 5) (s - 5)}\} \\ = & L^{- 1} \{- \frac{9}{25 s^{2}} - \frac{9}{250 (s + 5)} + \frac{9}{250 (s - 5)}\} \\ = & \frac{- 9}{25} L^{- 1} \{\frac{1}{s^{2}}\} - \frac{9}{250} L^{- 1} \{\frac{1}{(s + 5)}\} + \frac{9}{250} L^{- 1} \{\frac{1}{s - 5}\} \\ X (t) & = & \frac{- 9 t}{25} - \frac{9}{250} e^{- 5 t} + \frac{9}{250} e^{5 t} \end{array}$

Hence solution of the IVP is $\begin{array}{rcl} X (t) = \frac{9}{250} e^{5 t} - \frac{9}{250} e^{- 5 t} - \frac{9 t}{25} \end{array}$

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