Laplace Transform with Initial Value Problems (IVPs)

Problem 129: Consider the initial value problem

(1) $\begin{equation*} y'' + 9y = 36t, \qquad y(0) = 7, \quad y'(0) = 2.\end{equation*}$

(a) Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of $y(t)$ by $Y(s)$ .

(b) Solve your equation for $Y(s)$ .

Solution: First, let’s highlights some of the things we need to solve this problem. That is,

(2) $\begin{align*} \mathcal{L} \{y''\} & = s^2 \mathcal{L}\{y\} - sy(0) - y'(0), \\ \mathcal{L}\{y\} & = Y(s), \text{and} \\ \mathcal{L}\{36 t\} & = \int_{0}^{\infty} 36t \cdot e^{-2t} \ dt = \frac{36}{s^2}.\end{align*}$

(a) Let’s now find the corresponding Laplace transform.

(3) $\begin{align*} y'' + 9y & = 36t \\ \mathcal{L}\{y''\} + \mathcal{L}\{9y\} & = \mathcal{L}\{36t\} \\ \mathcal{L}\{y''\} + 9\mathcal{L}\{y\} & = 36\mathcal{L}\{t\} \\ s^2\mathcal{L}\{y\} - sy(0) - y'(0) + 9 \mathcal{L}\{y\} & = \frac{36}{s^2} \\ s^2 Y(s) - 7s - 2 + 9 Y(s) & = \frac{36}{s^2} \\ (s^2+9)Y(s) - 7s - 2 & = \frac{36}{s^2}.\end{align*}$

(b) One can now some for $Y(s)$ .

(4) $\begin{align*} (s^2+9)Y(s) - 7s - 2 & = \frac{36}{s^2} \\ (s^2 + 9) Y(s) & = \frac{36}{s^2} + 7s + 2 \\Y(s) & = \frac{36}{s^2 \cdot (s^2+9)} + \frac{7s}{s^2+9} + \frac{2}{s^2+9} \\ \mathcal{L}\{y(t)\} & = \frac{36}{s^2 \cdot (s^2+9)} + \frac{7s}{s^2+9} + \frac{2}{s^2+9} .\end{align*}$

That is, we can now find the solution $y(t)$ by taking the inverse Laplace transform.

(5) $\begin{equation*} y(t) = \mathcal{L}^{-1}\bigg\{\frac{36}{s^2 \cdot (s^2+9)} + \frac{7s}{s^2+9} + \frac{2}{s^2+9} \bigg\}.\end{equation*}$

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Laplace Transform with Initial Value Problems (IVPs)

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