DE-BOOK Step 2 — Solving in the Laplace Domain

Section Step 2 — Solving in the Laplace Domain

Once we’ve applied the Laplace transform to a differential equation, we’re left with a new algebraic equation involving \(Y(s)\text{.}\) Solving this equation is usually easier than solving the original differential equation, but we still need to work carefully. In this section, you’ll learn how to isolate \(Y(s)\text{,}\) clean it up using algebra, and get it into a form that matches the inverse transform table.

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Subsection What It Means to Solve in the Laplace Domain

After transforming a differential equation into the Laplace domain, you’re left with an algebraic equation involving \(Y(s)\text{.}\) We will call this the Laplace-domain equation. In this setting, your new unknown is now represented by its Laplace transform, \(Y(s)\text{.}\)

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Your goal is twofold: first, solve for \(Y(s)\) using algebra, and second, rewrite \(Y(s)\) so it matches known inverse transform forms. That prepares it for the final step of returning to the original domain.

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Step 2️⃣: Solve for \(Y\) & Prepare it for an Inverse Transform.

\begin{equation*} \DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{align*} \amp \DLBa s^2Y - 9Y - s + 7 = \frac{10}{s - 2} \amp\amp\!\!\!\!\!\!\!\!\DLBa ◀\ \ \begin{array}{l}\text{Laplace Domain}\\ \text{Equation}\end{array}\\ \amp\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp \DLBa Y = \frac{1}{s^2 - 9}\left( \frac{10}{s - 2} + s - 7 \right) \amp\amp\!\!\!\!\!\!\!\!\DLBa ◀\ \ \text{Isolated}\ Y \\ \amp\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp \DLBa Y = \frac{1}{s - 3} + \frac{2}{s + 3} - \frac{2}{s - 2} \amp\amp\!\!\!\!\!\!\!\!\DLBa ◀\ \ \begin{array}{l}\text{Ready for}\\ \text{Inverse}\end{array} \end{align*}

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Checkpoint 208. 📖❓ What are we solving for?

What variable do you solve for in the Laplace-domain equation?

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\(\ Y\text{,}\) the Laplace transform of \(y\)
Exactly! You’re solving for \(Y(s)\text{,}\) which represents the Laplace transform of \(y(t)\text{.}\)
\(\ y\text{,}\) the dependent variable of the differential equation
Not yet. You’ll get \(y(t)\) by inverting \(Y(s)\) later.
\(\ s\text{,}\) the Laplace-domain variable
Nope. \(s\) is a variable, not the unknown you’re solving for.
\(\ t\text{,}\) the independent variable of the differential equation
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Subsection Solving for \(Y(s)\)

The first step after transforming your differential equation into a Laplace-domain equation is to solve for \(Y(s)\text{.}\) This means using standard algebra to isolate \(Y(s)\) on one side of the equation. Here is a quick example.

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🌌 Example 209. Solving a Laplace-Domain Equation.

Solve the Laplace-domain equation for \(Y(s)\text{:}\)

\begin{equation*} s^2 Y - 2s - 4sY + 5 + 6Y = \frac{2}{s^3}. \end{equation*}

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Solution.

Start by moving all non-\(Y\) terms to the right-hand side:

\begin{equation*} s^2Y - 4sY + 6Y = \frac{2}{s^3} + 2s - 5. \end{equation*}

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Now, factor \(Y\) from the remaining terms on the left-hand side:

\begin{equation*} Y(s^2 - 4s + 6) = \frac{2}{s^3} + 2s - 5. \end{equation*}

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Finally, divide both sides by the quadratic expression to isolate \(Y(s)\text{:}\)

\begin{equation*} Y(s) = \frac{1}{s^2 - 4s + 6} \left( \frac{2}{s^3} + 2s - 5 \right). \end{equation*}

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Subsection Preparing for the Inverse Transform

Once you’ve solved for \(Y(s)\text{,}\) the next step is to prepare it for returning to the original domain. This means rewriting \(Y(s)\) as a combination of expressions that match known Laplace transforms, so we can use the inverse transform to find \(y(t)\text{.}\)

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The inverse Laplace transform works by reversing the process of the forward transform. This can be done using the same table of transforms—but in reverse. Our goal is to express \(Y(s)\) as a sum of terms that match entries in the \(s\)-domain column.

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Preparation Guidelines.

After isolating \(Y\) from the transformed equation, you typically are left with a messy combination of rational functions in the variable \(s\text{.}\)

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Since we can only invert functions in the \(s\)-domain column of the table of common Laplace transforms, it helps to list them out to guide us on how to prepare \(Y\text{.}\)

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\begin{equation} \frac{1}{s} \end{equation}	\begin{equation} \frac{n!}{s^{n+1}} \end{equation}	\begin{equation} \frac{s}{s^2+b^2} \end{equation}	\begin{equation} \frac{b}{s^2+b^2} \end{equation}
\begin{equation} \frac{1}{s - a} \end{equation}	\begin{equation} \frac{n!}{(s-a)^{n+1}} \end{equation}	\begin{equation} \frac{s-a}{(s-a)^2+b^2} \end{equation}	\begin{equation} \frac{b}{(s-a)^2+b^2} \end{equation}

To invert \(Y\text{,}\) it must first be written as the sum of one or more of these terms. However, since the first row is a special case of the second when \(a=0\text{,}\) we can reduce the list down to the following function types:

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\begin{equation} \frac{1}{s - a},\quad \frac{1}{(s - a)^{n+1}},\quad \frac{b}{(s-a)^2+b^2},\quad \frac{s-a}{(s-a)^2+b^2},\tag{45} \end{equation}

Fortunately, there is a technique that was designed for this exact task. Recall that partial fraction decomposition (PFD) guarantees that any rational function can be written as the sum of rational terms with linear or quadratic factors in the denominator.

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Therefore, if the denominator of \(Y\) has degree three or higher, PFD will always be your first step toward preparing \(Y\) for a backward transform out of the Laplace domain.

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Dealing with Quadratic Denominators.

After partial fraction decomposition, \(Y\) will be the sum of rational terms with either a linear or quadratic factor in the denominator. If they are all linear then you are done and you can move on to inverting \(Y\text{.}\) In contrast, if at least one term has a quadratic denominator then it will be of the form

\begin{equation*} \frac{\text{*}}{As^2 + Bs + C}\text{.} \end{equation*}

where we have omitted the numerator since will be not be important to this discussion.

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Notice that the denominator is not nicely factored like you see in the transform table. To achieve this, we need to complete the square.

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Note that if the quadratic denominator term was the result of partial fraction decomposition, then it cannot be reduced further and you should skip directly to completing the square of the denominator. However, if \(Y\) started with this form, then you should first check to see if the denominator is factorable. If so, then use PFD. Otherwise, complete the square of the denominator so that it resembles the form in the table:

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\begin{equation*} \frac{*}{As^2 + Bs + C}\quad\rightarrow\quad\frac{*}{(s - a)^2 + b^2}, \end{equation*}

Checkpoint 210. 📖❓ Complete the Square.

Give the correct form after completing the square of the denominator:

\begin{equation*} Y(s) = \frac{11}{s^2 - 6s + 14} = \fillinmath{XXXXX} \end{equation*}

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\(\ds\frac{11}{(s - 3)^2 + 5}\)
Correct
\(\ds\frac{11}{(s + 3)^2 + 9}\)
Incorrect
\(\ds\frac{11}{(s - 1)^2 + 14}\)
Incorrect
\(\ds\frac{11}{(s - 2)^2 + 10}\)
Incorrect

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Subsection Preparation Examples

In the following examples, we prepare \(Y\) for an inverse Laplace transform by writing it as the sum of the following types of terms:

\begin{equation*} \frac{1}{s - a},\quad \frac{1}{(s - a)^{n+1}},\quad \frac{b}{(s-a)^2+b^2},\quad \frac{s-a}{(s-a)^2+b^2} \text{.} \end{equation*}

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Partial Fraction Decomposition Examples.

🌌 Example 211. Distinct Linear Factors.

\begin{equation*} Y(s) = \frac{2s + 5}{s^2 + 5s + 4} \end{equation*}

Solution.

Since the denominator can be factored as \((s+1)(s+4)\text{,}\) we apply partial fraction decomposition:

\begin{equation*} \frac{2s+5}{(s+1)(s+4)} = \frac{A}{s+1} + \frac{B}{s+4} \end{equation*}

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To find \(A\) and \(B\text{,}\) we multiply both sides by the common denominator, \((s+1)(s+4)\text{,}\) eliminating all the fractions and leaving us with

\begin{equation*} 2s + 5 = A(s + 4) + B(s + 1)\text{.} \end{equation*}

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Strategically selecting \(s\) as \(-1\) and \(-4\) leads directly to \(A\) and \(B\text{:}\)

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\begin{align*} s = -1: \amp\\ \\ s = -4: \amp\\ \end{align*}

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\begin{align*} 2(-1) + 5 \amp = A(-1 + 4) + B(-1 + 1)\\ 3 \amp = A(3) + B(0)\\ 2(-4) + 5 \amp = A(-4 + 4) + B(-4 + 1)\\ -3 \amp = A(0) + B(-3) \end{align*}

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\begin{gather*} \\ A=1\\ \\ B=1 \end{gather*}

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This completes the preparation of \(Y\) for an inverse transform.

\begin{equation*} Y(s) = \frac{1}{s + 1} + \frac{1}{s + 4}\text{.} \end{equation*}

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🌌 Example 212. Repeated Linear Factor.

\begin{equation*} Y(s) = \frac{3s + 4}{s^2 - 2s + 1} \end{equation*}

Solution.

Factor the quadratic denominator:

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\begin{equation*} s^2 - 2s + 1 = (s - 1)^2\text{.} \end{equation*}

Decompose \(Y(s)\) into partial fractions:

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\begin{equation*} \frac{3s + 4}{(s - 1)^2} = \frac{A}{s - 1} + \frac{B}{(s - 1)^2}. \end{equation*}

Multiply both sides by \((s - 1)^2\text{,}\)

\begin{equation*} 3s + 4 = A(s - 1) + B. \end{equation*}

and solve for \(A\) and \(B\) by selecting convenient values for \(s\text{:}\)

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\begin{align*} s = 1: \amp\\ \\ s = 0: \amp\\ \end{align*}

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\begin{align*} 3(1) + 4 \amp = A(1 - 1) + B\\ 7 \amp = A(0) + B \quad\implies\quad B=7\\ 3(0) + 4 \amp = A(0 - 1) + B\\ 4 \amp = -A + 7 \quad\implies\quad A=3 \end{align*}

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The prepared form of \(Y(s)\) is

\begin{equation*} Y(s) = \frac{3}{s - 1} + \frac{7}{(s - 1)^2}. \end{equation*}

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🌌 Example 213. Three Linear Factors.

\begin{equation*} Y(s) = \frac{2s^2 + 5s + 1}{s^3 - s} \end{equation*}

Solution.

Since the degree of the denominator is greater than \(2\text{,}\) we use PFD and we factor the denominator as

\begin{equation*} s^3 - s = s(s - 1)(s + 1) \end{equation*}

so that \(Y(s)\) has the form

\begin{equation*} Y(s) = \frac{2s^2 + 5s + 1}{s(s - 1)(s + 1)} = \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s + 1}\text{.} \end{equation*}

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Multiply both sides by \(s(s - 1)(s + 1)\) to get

\begin{equation*} 2s^2 + 5s + 1 = A(s - 1)(s + 1) + B(s)(s + 1) + C(s)(s - 1) \end{equation*}

and choose convenient values for \(s\) so that

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\begin{align*} s = 0: \amp \\ \\ s = 1: \amp \\ \\ s = -1: \amp \\ \end{align*}

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\begin{align*} 1 \amp = A(-1)(1) + B(0)(1) + C(0)(-1) \\ 1 \amp = -A \\ 8 \amp = A(0)(2) + B(1)(2) + C(1)(0) \\ \amp = 2B \\ -2 \amp = A(1)(0) + B(-1)(0) + C(-1)(-2) \\ -2 \amp = 2C \end{align*}

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\begin{align*} \\ A \amp = -1 \\ \\ B \amp = 4 \\ \\ C \amp = -1 \end{align*}

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Therefore, the prepared form of \(Y(s)\) is

\begin{equation*} Y(s) = \frac{-1}{s} + \frac{4}{s - 1} + \frac{-1}{s + 1} \end{equation*}

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Checkpoint 214. 📖❓ Partial Fraction Decomposition.

Complete the partial fraction decomposition:

\begin{equation*} Y(s) = \frac{2s + 3}{(s + 2)(s + 3)} = \end{equation*}

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\(\ \dfrac{-1}{s+2} + \dfrac{3}{s+3}\)
Correct!
\(\ \dfrac{1}{s+2} + \dfrac{-3}{s+3}\)
Incorrect
\(\ \dfrac{-1}{s+2} + \dfrac{1}{s+3}\)
Incorrect
\(\ \dfrac{2}{s+2} + \dfrac{3}{s+3}\)
Incorrect

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Completing the Square Examples.

🌌 Example 215. Completing the Square Example.

\begin{equation*} Y(s) = \frac{s-1}{s^2 - 2s + 5} \end{equation*}

Solution.

A quick calculation of the discriminant shows the denominator is not factorable:

\begin{equation*} \Delta = b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 \lt 0\text{.} \end{equation*}

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So we complete the square of the denominator:

\begin{align*} s^2 - 2s + 5 \amp = \ob{s^2 - 2s + \boxed{1}}^{(s - 1)^2} - \boxed{1} + 5 \\ \amp = (s - 1)^2 + 4 \text{.} \end{align*}

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Substituting this into \(Y(s)\) gives the prepared form

\begin{equation*} Y(s) = \frac{s - 1}{s^2 - 2s + 5} = \frac{s - 1}{(s - 1)^2 + 4}\text{,} \end{equation*}

which is a perfect match with L\(_6\) with \(a=1\text{.}\)

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Mixed Technique Examples.

🌌 Example 216. Completing the Square Example.

\begin{equation*} Y(s) = \frac{4s^3 + 3s^2 - 2s + 1}{s^4 - 1} \end{equation*}

Solution.

Since the degree of the denominator is greater than \(2\text{,}\) we use PFD and factor the denominator:

\begin{equation*} s^4 - 1 = (s^2 - 1)(s^2 + 1) = (s - 1)(s + 1)(s^2 + 1)\text{.} \end{equation*}

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Decompose \(Y(s)\) into partial fractions:

\begin{equation*} \frac{4s^3 + 3s^2 - 2s + 1}{(s - 1)(s + 1)(s^2 + 1)} = \frac{A}{s - 1} + \frac{B}{s + 1} + \frac{Cs + D}{s^2 + 1} \end{equation*}

and multiply both sides by \((s - 1)(s + 1)(s^2 + 1)\) to get

\begin{align*} 4s^3 + 3s^2 \amp - 2s + 1 \\ \amp = A(s + 1)(s^2 + 1) + B(s - 1)(s^2 + 1) + (Cs+D)(s - 1)(s + 1) \text{.} \end{align*}

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Strategically selecting values for \(s\text{,}\) we get

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\(s=1:\)

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\(6 = A(2)(2) + B(0)(2) + (C(1) + D)(0)(2)\)

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\(\vphantom{\dfrac11}\quad\Rightarrow\quad\boxed{A = \sfrac32}\)

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\(s=-1:\)

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\(2 = A(0)(2) + B(-2)(2) + (C(-1) + D)(-2)(0)\)

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\(\vphantom{\dfrac11}\quad\Rightarrow\quad\boxed{B = -\sfrac12}\)

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\(s=0:\)

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\(1 = A(1)(1) + B(-1)(1) + (C(0) + D)(-1)(1)\)

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\(\vphantom{\dfrac11}\quad\Rightarrow\quad\boxed{D = 1}\)

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\(s=3:\)

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\(-15 = A(-1)(3) + B(-3)(3) + (C(-2) + D)(-3)(-1)\)

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\(\vphantom{\dfrac11}\quad\Rightarrow\quad\boxed{C = 3}\)

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Therefore, the prepared Laplace domain solution is

\begin{equation*} Y(s) = \frac{\sfrac32}{s - 1} - \frac{\sfrac12}{s + 1} + \frac{3s + 1}{s^2 + 1}\text{.} \end{equation*}

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🌌 Example 217. Completing the Square Example.

\begin{equation*} Y(s) = \frac{3s + 4}{(s^2 + 1)(s^2 + 4)} \end{equation*}

Solution.

Partial fraction decomposition gives:

\begin{equation*} Y(s) = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 4}\text{.} \end{equation*}

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Multiply both sides by \((s^2 + 1)(s^2 + 4)\) to get

\begin{equation*} 3s + 4 = (As + B)(s^2 + 4) + (Cs + D)(s^2 + 1)\text{.} \end{equation*}

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When you have multiple quadratic terms involved, it is usually easier to match coefficients rather than choosing values of \(s\text{.}\) So let’s multiply out the right and match the corresponding terms to find \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\text{:}\)

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\begin{align*} 3s + 4 \amp = (As + B)(s^2 + 4) + (Cs + D)(s^2 + 1) \\ \amp = As^3 + Bs^2 + 4As + 4B + Cs^3 + Ds^2 + Cs + D \\ \amp = (A + C)s^3 + (B + D)s^2 + (4A + C)s + (4B + D) \end{align*}

Matching the coefficients leads to the following system:

\begin{equation*} A + C = 0,\qquad B + D = 0,\qquad 4A + C = 3,\qquad 4B + D = 4 \end{equation*}

which has the solution: \(A = 1\text{,}\) \(B = \sfrac43\text{,}\) \(C = -1\) and \(D = -\sfrac43\text{.}\)

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Therefore, the prepared Laplace domain solution is given by:

\begin{equation*} Y(s) = \frac{s + \sfrac43}{s^2 + 1} - \frac{s - \sfrac43}{s^2 + 4}\text{.} \end{equation*}

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Checkpoint 218. 📖❓ Preparing \(Y(s)\).

What should be done to prepare \(Y(s)\) for an inverse?

\begin{equation*} Y(s) = \frac{s + 6}{s^2 + 7s + 6}\text{?} \end{equation*}

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Complete the square of the denominator
Correct!
Partial fraction decomposition
Is the denominator factorable?
Cancel out the \(6\) in the numerator and denominator.
Canceling out the \(6\) in the function is not a valid operation.
Cancel out the \(s\) in the numerator and denominator.
Canceling out the \(s\) in the function is not a valid operation.

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Step 2 — Solving in the Laplace Domain Summary.

Treat the transformed equation as pure algebra: isolate \(Y(s)\text{,}\) then reshape it so every term matches an entry in the inverse-transform table.

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

Once you have a Laplace-domain equation, your goal is to isolate \(Y(s)\text{.}\)

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After solving for \(Y(s)\text{,}\) you need to rewrite it in a form that matches known inverse transforms.

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Partial fraction decomposition helps split complicated rational expressions into easier pieces.

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Quadratic terms in the denominator often require completing the square.

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You have attempted of activities on this page.

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Prev Top Next

\begin{equation} \frac{1}{s} \end{equation}	\begin{equation} \frac{n!}{s^{n+1}} \end{equation}	\begin{equation} \frac{s}{s^2+b^2} \end{equation}	\begin{equation} \frac{b}{s^2+b^2} \end{equation}
\begin{equation} \frac{1}{s - a} \end{equation}	\begin{equation} \frac{n!}{(s-a)^{n+1}} \end{equation}	\begin{equation} \frac{s-a}{(s-a)^2+b^2} \end{equation}	\begin{equation} \frac{b}{(s-a)^2+b^2} \end{equation}