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Chapter 8: Problem 14

Use the Laplace transform to solve the initial value problem. \(y^{\prime \prime}-y^{\prime}-6 y=2, \quad y(0)=1, \quad y^{\prime}(0)=0\)

Short Answer

Expert verified
Question: Using the Laplace transform, find the solution to the initial value problem \(y''(t) - y'(t) -6y(t) = 2\) with \(y(0) = 1\) and \(y'(0) = 0\). Answer: The solution to the initial value problem is \(y(t) = -1 + \frac{1}{5} e^{3t} \cos(\sqrt{8} t) + \frac{7}{5} e^{3t} \sin(\sqrt{8} t)\).

Step by step solution

01

Apply Laplace Transform to the given equation

The given equation is \(y''(t) - y'(t) - 6y(t) = 2\). To transform the derivatives, we will use the following properties: - \(L \{y'(t)\} = sY(s) - y(0)\) - \(L \{y''(t)\} = s^2Y(s) - sy(0) - y'(0)\) Now, applying the Laplace transform to the given equation: \begin{align*} L\{y''(t) - y'(t) - 6y(t)\} &= L\{2\} \\ s^2Y(s) - sy(0) - y'(0) - sY(s) + y(0) - 6Y(s) &= \frac{2}{s} \end{align*}
02

Substitute initial conditions

Now, let's substitute the initial conditions \(y(0)=1\) and \(y'(0)=0\): \begin{align*} s^2Y(s) - s - 0 - sY(s) + 1 - 6Y(s) &= \frac{2}{s} \\ (s^2 - s + 1 - 6s)Y(s) &= \frac{2}{s} - 1 \end{align*}
03

Solve for Y(s)

Next, we will solve the equation for \(Y(s)\): \begin{align*} (6s - s^2 + 1)Y(s) &= \frac{2}{s} - 1 \\ Y(s) &= \frac{(2 - s)}{s(s^2 - 6s + 1)} \end{align*}
04

Apply Inverse Laplace Transform

Now, we will apply the inverse Laplace transform to find y(t): $$ y(t) = L^{-1}\{Y(s)\} = L^{-1}\left\{\frac{(2 - s)}{s(s^2 - 6s + 1)}\right\} $$
05

Partial fraction decomposition and inverse Laplace transform

Before applying inverse Laplace transform, we will have to perform partial fraction decomposition to find \(A\), \(B\), and \(C\). Let: $$ \frac{(2 - s)}{s(s^2 - 6s + 1)} = \frac{A}{s} + \frac{Bs + C}{s^2 - 6s + 1} $$ Solving for \(A\), \(B\), and \(C\), we get \(A=-1\), \(B=\frac{1}{5}\), and \(C=\frac{7}{5}\). Now we can write the inverse Laplace transform as: $$ y(t) = -1 + \frac{1}{5} e^{3t} \cos(\sqrt{8}t) + \frac{7}{5} e^{3t} \sin(\sqrt{8}t) $$ Therefore, the solution to the initial value problem is: $$ y(t) = -1 + \frac{1}{5} e^{3t} \cos(\sqrt{8}t) + \frac{7}{5} e^{3t} \sin(\sqrt{8}t) $$

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Most popular questions from this chapter

In Exercises 26-28, solve the initial value problem $$ \begin{array}{|l|l|l|} \hline \mathrm{L} & y^{\prime \prime}+3 y^{\prime}+2 y=f_{h}(t), \quad y(0)=0, & y^{\prime}(0)=0 \end{array} $$

In Exercises 26-28, solve the initial value problem $$ \mathrm{L} \quad y^{\prime \prime}+2 y^{\prime}+2 y=f_{h}(t), \quad y(0)=0, \quad y^{\prime}(0)=0 $$

Express the given function \(f\) in terms of unit step functions and use Theorem 8.4 .1 to find \(\mathcal{L}(f) .\) Where indicated by, graph \(f\). $$ (t)=\left\\{\begin{array}{cc} -t, & 0 \leq t<2 ,\\\ t-4, & 2 \leq t<3 ,\\\ 1, & t \geq 3. \end{array}\right. $$

In Exercises \(1-20\) solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ y^{\prime \prime}+4 y=8 e^{2 t}+\delta(t-\pi / 2), \quad y(0)=8, \quad y^{\prime}(0)=0 $$

Use the Laplace transform to solve the initial value problem. \(y^{\prime \prime}+2 y^{\prime}+2 y=2 t, \quad y(0)=2, \quad y^{\prime}(0)=-7\)
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