MHB Collin's question via email about solving a DE using Laplace Transforms

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The discussion focuses on solving the differential equation y'' + 4y = H(t - 7) with initial conditions y(0) = 0 and y'(0) = -10 using Laplace Transforms. The Laplace Transform is applied to both sides, leading to an expression for Y(s) that incorporates the Heaviside function and initial conditions. Partial fraction decomposition is used to simplify the inverse transform process, allowing the solution to be expressed in terms of sine functions and Heaviside functions. The final solution is y(t) = H(t - 7) - (1/8)sin[2(t - 7)]H(t - 7) - 5sin(2t), which is verified to satisfy the original differential equation. The method is noted for its efficiency and reduced potential for error compared to alternative approaches.
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Solve the following Differential Equation:

$\displaystyle \begin{align*} y'' + 4\,y = \mathrm{H}\,\left( t - 7 \right) \textrm{ with } y\left( 0 \right) = 0 \textrm{ and } y'\left( 0 \right) = -10 \end{align*}$

Taking the Laplace Transform of both sides we have

$\displaystyle \begin{align*} \mathcal{L}\,\left\{ y'' + 4\,y \right\} &= \mathcal{L}\,\left\{ \mathrm{H}\,\left( t - 7 \right) \right\} \\ s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &= \frac{\mathrm{e}^{-7\,s}}{s} \\ s^2\,Y\left( s \right) + 10 + 4\,Y\left( s \right) &= \frac{ \mathrm{e}^{-7\,s}}{s} \\ \left( s^2 + 4 \right) \, Y\left( s \right) &= \frac{\mathrm{e}^{-7\,s}}{s} - 10 \\ Y\left( s \right) &= \frac{\mathrm{e}^{-7\,s}}{s\,\left( s^2 + 4 \right) } - \frac{10}{s^2 + 4} \end{align*}$

So to solve the DE, all that we require now is to take the Inverse Laplace Transform of this result. To do this with the first term, we will need to use Partial Fractions:

$\displaystyle \begin{align*} \frac{A}{s} + \frac{B\,s + C}{s^2 + 4} &\equiv \frac{1}{s\,\left( s^2 + 4 \right) } \\ A\,\left( s^2 + 4 \right) + \left( B\,s + C \right) \, s &= 1 \end{align*}$

Let $\displaystyle \begin{align*} s = 0 \end{align*}$ to find $\displaystyle \begin{align*} 4\,A = 1 \implies A = \frac{1}{4} \end{align*}$.

The coefficient of $\displaystyle \begin{align*} s^2 \end{align*}$ is $\displaystyle \begin{align*} A + B \end{align*}$ on the left and $\displaystyle \begin{align*} 0 \end{align*}$ on the right, and as $\displaystyle \begin{align*} A = \frac{1}{4} \end{align*}$ we have $\displaystyle \begin{align*} \frac{1}{4} + B = 0 \implies B = -\frac{1}{4} \end{align*}$.

The coefficient of $\displaystyle \begin{align*} s \end{align*}$ is $\displaystyle \begin{align*} C \end{align*}$ on the left and $\displaystyle \begin{align*} 0 \end{align*}$ on the right, thus $\displaystyle \begin{align*} C = 0 \end{align*}$.

Therefore $\displaystyle \begin{align*} \frac{1}{s\,\left( s^2 + 4 \right) } = \frac{1}{4\,s} - \frac{s}{4\,\left( s^2 + 4 \right) } \end{align*}$. So that means

$\displaystyle \begin{align*} y &= \mathcal{L}^{-1}\,\left\{ \frac{\mathrm{e}^{-7\,s}}{s\,\left( s^2 + 4 \right) } - \frac{10}{s^2 + 4} \right\} \\ &= \frac{1}{4} \mathcal{L}^{-1}\,\left\{ \frac{\mathrm{e}^{-7\,s}}{s} \right\} - \frac{1}{8} \,\mathcal{L}^{-1}\,\left\{ \mathrm{e}^{-7\,s} \,\left( \frac{2}{ s^2 + 2^2 } \right) \right\} - 5\,\mathcal{L}^{-1}\,\left\{ \frac{2}{s^2 + 2^2} \right\} \end{align*}$

The first and third terms are very straightforward. For the second term, we need to make use of the rule $\displaystyle \begin{align*} \mathcal{L}\,\left\{ f\left( t - c \right) \, \mathrm{H}\,\left( t - c \right) \right\} = \mathrm{e}^{-c\,s}\,F\left( s \right) \end{align*}$. We can read off that $\displaystyle \begin{align*} F\left( s \right) = \frac{2}{ s^2 + 2^2} \end{align*}$ and thus $\displaystyle \begin{align*} f\left( t \right) = \sin{ \left( 2\,t \right) } \end{align*}$. From here we can get that $\displaystyle \begin{align*} f\left( t - 7 \right) = \sin{ \left[ 2\,\left( t - 7 \right) \right] } \end{align*}$, and finally the solution to the DE is

$\displaystyle \begin{align*} y = \mathrm{H}\,\left( t - 7 \right) - \frac{1}{8}\sin{ \left[ 2\,\left( t - 7 \right) \right] }\,\mathrm{H}\,\left( t - 7 \right) - 5 \sin{ \left( 2\,t \right) } \end{align*}$.
 
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Careful students will check the solution to ensure that it actually is a solution to the DE.
 
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