DE-BOOK Step-by-Step Examples

Section 11.2 Step-by-Step Examples

Now that we’ve explored the theory behind the Laplace Transform Method, it’s time to put it into action by solving some initial-value problems. We’ll walk through two examples that will illustrate each of the four key steps in the method. Along the way, we’ll highlight how the forward and backward Laplace transforms work together to convert a differential equation into an algebraic problem and then back to a solution. Let’s get started!

Example 24. Solve the initial-value problem.

\begin{matrix} (48) & y^{'} + y = 4, y (0) = 2 . \end{matrix}

Solution 1. Step 1: Apply the Forward Transform

Step 1. Applying

L

to both sides of (48), we have

\begin{matrix} L {y^{'} + y} = L {4} \\ L {y^{'}} + L {y} = L {4} \\ s Y (s) - 2 + Y (s) = \frac{4}{s} \end{matrix}

Solution 2. Step 2a: Solve for

Y (s)

Step 2. Next, we isolate

Y (s)

on one side of the equation:

\begin{aligned} s Y (s) - 2 + Y (s) = & \frac{4}{s} \\ (s + 1) Y (s) = & \frac{4}{s} + 2 \\ Y (s) = & \frac{1}{s + 1} (\frac{4}{s} + 2) \\ Y (s) = & \underset{(*)}{\underset{⏟}{\frac{4}{s (s + 1)}}} + \underset{(* *)}{\underset{⏟}{\frac{2}{s + 1}}} \end{aligned}

Solution 3. Step 2b: Prepare for the Backward Transform

Step 3. We need to express

Y (s)

as a sum of functions that match known forms in the Laplace transform table. We see from the last equation,

(* *)

is ready to go, but

(*)

requires partial fraction decomposition. We start by writing down the form of the decomposition,

(*) = \frac{4}{s (s + 1)} = \frac{A}{s} + \frac{B}{s + 1} or

\begin{matrix} (49) & 4 = A (s + 1) + B s \end{matrix}

Now, we find

A

and

B

by plugging in values of

s

into (49),

$s = 0$ :	$4$	$=$	$A (0 + 1) + B (0)$
	$4$	$=$	$A$
$s = - 1$ :	$4$	$=$	$A (- 1 + 1) + B (- 1)$
	$- 4$	$=$	$B$

Thus, the prepared

Y (s)

Y (s) = \frac{4}{s} + \frac{- 4}{s + 1}

Solution 4. Step 3: Apply the Backward Transform

Step 4. Finally, we apply the backward step to get the solution:

\begin{aligned} L^{- 1} {Y (s)} = & L^{- 1} {\frac{4}{s} + \frac{- 4}{s + 1}} \\ y (t) = & 4 - 4 e^{- t} \end{aligned}

Example 25. Solve the initial-value problem.

\begin{matrix} (50) & y^{″} - 3 y^{'} + 2 y = e^{2 t}, y (0) = 1, y^{'} (0) = 0 \end{matrix}

Solution 1. Step 1: Apply the Forward Transform

Step 1. Applying

L

to both sides of (50), we have

\begin{aligned} L {y^{″} - 3 y^{'} + 2 y} = & L {e^{2 t}} \\ L {y^{″}} - 3 L {y^{'}} + 2 L {y} = & \frac{1}{s - 2} \end{aligned}

where

L {y (t)} = Y (s)

and

$L {y^{'}} = s Y (s) - y (0) = s Y (s) - 1,$
$L {y^{″}} = s^{2} Y (s) - s y (0) - y^{'} (0) = s^{2} Y (s) - s .$

So the complete forward transform of (50) is

s^{2} Y (s) - s - 3 [s Y (s) - 1] + 2 Y (s) = \frac{1}{s - 2}

Solution 2. Step 2a: Solve for

Y (s)

Step 2. Next, we isolate

Y (s)

on one side of the equation:

\begin{matrix} s^{2} Y (s) - s - 3 s Y (s) + 3 + 2 Y (s) = \frac{1}{s - 2} \\ [s^{2} - 3 s + 2] Y (s) - s + 3 = \frac{1}{s - 2} \\ Y (s) [(s - 1) (s - 2)] = \frac{1}{s - 2} + s - 3 \\ Y (s) = \frac{1}{(s - 1) (s - 2)} \underset{(*)}{\underset{⏟}{(\frac{1}{s - 2} + s - 3)}} \end{matrix}

Solution 3. Step 2b: Prepare for the Backward Transform

Step 3. Now, we need to express

Y (s)

as a sum of functions that match known forms in the Laplace transform table. This can be simplified slightly by first combining the terms in

(*)

as a single fraction, like so

(*) = \frac{1}{s - 2} + \frac{(s - 3) (s - 2)}{s - 2} = \frac{1 + s^{2} - 5 s + 6}{s - 2} = \frac{s^{2} - 5 s + 7}{s - 2} .

Plugging this back into the equation for

(*)

gives the new

Y (s),

Y (s) = \frac{1}{(s - 1) (s - 2)} \cdot \frac{s^{2} - 5 s + 7}{s - 2} = \frac{s^{2} - 5 s + 7}{(s - 1) (s - 2)^{2}}

and we are ready to apply partial fraction decomposition. The form of the decomposition is

\begin{aligned} \frac{s^{2} - 5 s + 7}{(s - 1) (s - 2)^{2}} = & \frac{A}{s - 1} + \frac{B}{s - 2} + \frac{C}{(s - 2)^{2}} or \\ s^{2} - 5 s + 7 = & A (s - 2)^{2} + B (s - 1) (s - 2) + C (s - 1) . \end{aligned}

Now, we find

A, B,

and

C

by plugging in values of

s,

$s = 1$ :	$1 - 5 + 7$	$=$	$A (1 - 2)^{2} + B (1 - 1) (1 - 2) + C (1 - 1)$
	$3$	$=$	$A$
$s = 2$ :	$4 - 10 + 7$	$=$	$3 (2 - 2)^{2} + B (2 - 1) (2 - 2) + C (2 - 1)$
	$1$	$=$	$C$
$s = 0$ :	$7$	$=$	$3 (0 - 2)^{2} + B (0 - 1) (0 - 2) + 1 (0 - 1)$
	$7$	$=$	$12 + 2 B - 1$
	$- 2$	$=$	$B$

Thus, the prepared

Y (s)

Y (s) = \frac{3}{s - 1} + \frac{- 2}{s - 2} + \frac{1}{(s - 2)^{2}}

Solution 4. Step 3: Apply the Backward Transform

Step 4. Finally, we perform the backward step to get the solution:

\begin{aligned} L^{- 1} {Y (s)} \\ = & L^{- 1} {\frac{3}{s - 1}} + L^{- 1} {\frac{- 2}{s - 2}} + L^{- 1} {\frac{1}{(s - 2)^{2}}} \\ y (t) = & 3 e^{t} - 2 e^{2 t} + t e^{2 t} \end{aligned}

Example 26. Solve the initial-value problem.

\begin{matrix} (51) & x^{″} - 4 x^{'} + 13 x = 54 e^{- t}, x (0) = 0, x^{'} (0) = 0 \end{matrix}

Solution 1. Step 1: Apply the Forward Transform

Step 1. Applying

L

to both sides of (51), we have

\begin{matrix} L {x^{″} - 4 x^{'} + 13 x} = L {54 e^{- t}} \\ L {x^{″}} - 4 L {x^{'}} + 13 L {x} = \frac{54}{s + 1} \end{matrix}

where

L {x (t)} = X (s)

and

$L {x^{'}} = s X (s) - x (0) = s X (s),$
$L {x^{″}} = s^{2} X (s) - s x (0) - x^{'} (0) = s^{2} X (s) .$

So the complete forward transform of (51) is

s^{2} X (s) - 4 s X (s) + 13 X (s) = \frac{54}{s + 1}

Solution 2. Step 2a: Solve for

X (s)

Step 2. Next, we isolate

X (s)

on one side of the equation:

\begin{matrix} s^{2} X (s) - 4 s X (s) + 13 X (s) = \frac{54}{s + 1} \\ [s^{2} - 4 s + 13] X (s) = \frac{54}{s + 1} \\ X (s) = \frac{54}{(s + 1) (s^{2} - 4 s + 13)} \end{matrix}

Solution 3. Step 2b: Prepare for the Backward Transform

Step 3. Now, we need to express

X (s)

as a sum of functions that match known forms in the Laplace transform table, which requires partial fraction decomposition. We start by writing down the form of the decomposition,

\begin{matrix} \frac{54}{(s + 1) (s^{2} - 4 s + 13)} = \frac{A}{s + 1} + \frac{B s + C}{s^{2} - 4 s + 13} or \\ 54 = A (s^{2} - 4 s + 13) + (B s + C) (s + 1) \end{matrix}

Now, we find

A, B,

and

C

by plugging in values of

s,

$s = - 1$ :	$54$	$=$	$A (- 1)^{2} - 4 (- 1) + 13$
	$54$	$=$	$A + 4 + 13$
	$37$	$=$	$A$
$s = 0$ :	$54$	$=$	$37 (13) + (B (0) + C) (0 + 1)$
	$54$	$=$	$481 + C$
	$- 427$	$=$	$C$
$s = 1$ :	$54$	$=$	$37 (- 10) + (B (1) - 427) (1 + 1)$
	$54$	$=$	$- 370 + 2 B - 854$
	$1170$	$=$	$2 B$
	$585$	$=$	$B$

So the updated

X (s)

X (s) = \frac{37}{s + 1} + \frac{585 s - 427}{s^{2} - 4 s + 13} .

Note, the second term is not yet ready and we need to complete the square of its denominator before we can do the backward step.

\begin{aligned} X (s) = & \frac{37}{s + 1} + \frac{585 s - 427}{(s - 2)^{2} + 9} \\ = & \frac{37}{s + 1} + \frac{585 s - 2 (585) + 2 (585) - 427}{(s - 2)^{2} + 9} \\ = & \frac{37}{s + 1} + \frac{585 s - 2 (585)}{(s - 2)^{2} + 9} + \frac{2 (585) - 427}{(s - 2)^{2} + 9} \\ = & \frac{37}{s + 1} + 585 \frac{s - 2}{(s - 2)^{2} + 9} + 743 \frac{1}{(s - 2)^{2} + 9} \\ = & \frac{37}{s + 1} + 585 \frac{s - 2}{(s - 2)^{2} + 9} + \frac{743}{3} \frac{3}{(s - 2)^{2} + 9} \end{aligned}

Solution 4. Step 3: Apply the Backward Transform

Step 4. Now for the backward step to get the solution:

\begin{aligned} x (t) = & 37 e^{- t} + 585 \cos (3 t) + \frac{743}{3} \sin (3 t) \end{aligned}

These examples highlight how the Laplace Transform Method simplifies solving differential equations by converting them into algebraic equations and then back into the time domain. By mastering these steps, you can tackle a wide range of initial-value problems with ease and precision.

Reading Questions Check-Point Questions

1. Which step in this method accounts for the initial conditions?

Which step in this method accounts for the initial conditions?

Applying the Forward Transform
That’s correct! The initial conditions are accounted for when you apply the forward transform of the derivatives.
Solving for $Y (s)$
Incorrect. Please carefully review the Method.
Preparing for the Backward Transform
Incorrect. Please carefully review the Method.
Applying the Backward Transform
Incorrect. Please carefully review the Method.

2. It is possible to use both partial fraction decomposition and completing the square in the same problem using this method?

It is possible to use both partial fraction decomposition and completing the square in the same problem using this method?

True.
This is true! Both techniques are commonly used in the Laplace Transform Method to simplify the algebraic equation.
False.
This is true! Both techniques are commonly used in the Laplace Transform Method to simplify the algebraic equation.

3. Which mathematical technique would you not expect to see in Step 2b?

Which mathematical technique would you not expect to see in Step 2b?

Partial fraction decomposition
Incorrect. Partial fraction decomposition is commonly used in Step 2b.
Completing the square
Incorrect. Completing the square is often used in Step 2b.
Finding a common denominator.
Incorrect. This technique is also used in Step 2b.
Integration by Parts
Correct! Only algebra techniques are used in Step 2b.

4. The Laplace Transform Method can be used to solve nonhomogeneous differential equations.

The Laplace Transform Method can be used to solve nonhomogeneous differential equations

True
Correct! The Laplace Transform Method is especially useful for solving nonhomogeneous differential equations.
False
Incorrect. The method can indeed be used to solve nonhomogeneous differential equations.

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