The Art of Laplace Transforms
Solving IVPs

Have you ever wondered how to turn a complex differential equation into a simpler algebraic problem?

Think of the Laplace transform as a magical key that unlocks the world of differential equations, turning them into more manageable algebraic expressions.

This powerful tool helps us solve initial value problems by applying the Laplace transform to the derivatives involved.

Let’s take a closer look.

Laplace Transform of Derivatives

Key Formula

Laplace transforms of derivatives are ideally suited to solving linear initial value problems with constant coefficients, but why?

Let’s find out…

If we are given a function \(\mathrm{y}\), the transform of its derivative \(\mathrm{y}\) ‘ can be expressing in terms of:

\begin{equation}
L\left\{y^{\prime}\right\}=s L\{y\}-y(0)
\end{equation}

Which means, if \(f, f^{\prime}, \ldots, f^{(n-1)}\) are continuous on the interval \([0, \infty)\) and are of exponential order and if \(f^{(n)}(x)\) is piecewise continuous on \([0, \infty)\), then \(L\left\{f^{(n)}(x)\right\}=s^{n} F(s)-s^{n-1} f(0)-s^{n-2} f^{\prime}(0)-\cdots-f^{(n-1)}(0)\), where \(F(s)=L\{f(x)\}\).

In fact, this theorem shows how ideally suited the Laplace transform is for solving linear initial value problems in which the differential equation has constant coefficients.

Solving an Initial Value Problem Using Laplace Transform

Step-by-Step Example

This may seem strange at first, but let’s work through a question to see it in action.

Use the method of Laplace transforms to solve:

\begin{equation}
y^{\prime}-5 y=-e^{-2 t}, y(0)=3
\end{equation}

Step 1: First, we will take the Laplace transform of both sides of the differential equation:

\(L\left\{y^{\prime}-5 y\right\}=L\left\{-e^{-2 t}\right\}\)

Now we will use our operations and properties of Laplace transforms to transform the DE into an algebraic equation in terms of \(\mathrm{s}\) and solve for \(L\{y\}\).

Step 2: Apply Laplace transform to both sides of the equation:

\(L\left\{y^{\prime}\right\}-5 L\{y\}=L\left\{-e^{-2 t}\right\}\)

Step 3: Replace \(L\left\{y^{\prime}\right\}\) with \(s L(y)-y(0)\) and \(L\left\{-e^{-2 t}\right\}\) with \(\frac{-1}{s+2}\) using Laplace transform properties:

\(s L\{y\}-y(0)-5 L\{y\}=\frac{-1}{s+2}\)

Step 4: Combine the terms containing \(L\{y\}\) and replace \(y(0)\) with its initial value, 3:

\(L\{y\}(s-5)-3=\frac{-1}{s+2}\)

Step 5: Add 3 to both sides of the equation to isolate \(L\{y\}(s-5)\):

\(L\{y\}(s-5)=\frac{-1}{s+2}+3\)

Step 6: Combine the fractions on the right-hand side:

\(L\{y\}(s-5)=\frac{3 s+5}{s+2}\)

Step 7: Divide both sides by \((s-5)\) to isolate \(L\{y\}\) and obtain the Laplace transform of the solution:

\(L\{y\}=\frac{3 s+5}{(s+2)(s-5)}\)

Now all we have to do is apply the inverse Laplace transform to find the solution of the IVP with the help of Partial Fractions.

Step 8: Apply the inverse Laplace transform to find the solution:

\(y=L^{-1}\left\{\frac{3 s+5}{(s+2)(s-5)}\right\}\)

Step 9: Perform partial fraction decomposition:

\(\frac{3 s+5}{(s+2)(s-5)}=\frac{A}{(s+2)}+\frac{B}{(s-5)}\)

Step 10: Find the values of A and B:

\(\frac{3 s+5}{(s+2)(s-5)}=\frac{1 / 7}{(s+2)}+\frac{20 / 7}{(s-5)}\)

Step 11: Apply the inverse Laplace transform to each term separately:

\(y=L^{-1}\left\{\frac{1 / 7}{(s+2)}\right\}+L^{-1}\left\{\frac{20 / 7}{(s-5)}\right\}\)

Step 12: Factor out the constants from the inverse Laplace transforms:

\(y=\frac{1}{7} L^{-1}\left\{\frac{1}{s+2}\right\}+\frac{20}{7} L^{-1}\left\{\frac{1}{s-5}\right\}\)

Step 13: Finally, find the inverse Laplace transforms of the remaining terms to obtain the solution:

\(y=\frac{1}{7} e^{-2 t}+\frac{20}{7} e^{5 t}\)

And that’s it.

• Apply Laplace transform to both sides
• Use properties to transform the DE into an algebraic equation
• Solve the transformed equation \(L\{y\}\)
• Apply inverse Laplace transform for the solution of the original IVP

Don’t worry. Everything will make more sense once you see more examples.

Next Steps

Get ready, as you’ll…

• Learn how to take the derivative of a Laplace Transform and derive two essential formulas.
• Apply these formulas to solve several second-order differential equations.
• Discover how using Laplace Transforms for nonhomogeneous differential equations makes the process easier by allowing you to plug in your initial conditions as you go.

It’s going to be fun. So, let’s get to it!

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