- #1

Prove It

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MHB

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Solve the following IVP using Laplace Transforms:

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} + 11\,y = 3\,t, \quad y\left( 0 \right) = 5$

Take the Laplace Transform of the equation:

$\displaystyle \begin{align*} s\,Y\left( s \right) - y\left( 0 \right) + 11\,Y\left( s \right) &= \frac{3}{s^2} \\

s\,Y\left( s \right) - 5 + 11\,Y\left( s \right) &= \frac{3}{s^2} \\

\left( s + 11 \right) Y\left( s \right) &= \frac{3}{s^2} + 5 \\

Y\left( s \right) &= \frac{3}{s^2 \,\left( s + 11 \right) } + \frac{5}{s + 11} \end{align*}$

Apply Partial Fractions:

$\displaystyle \begin{align*}

\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + 11} &\equiv \frac{3}{s^2 \,\left( s + 11 \right) } \\

A\,s\left( s + 11 \right) + B \left( s + 11 \right) + C\,s^2 &\equiv 3

\end{align*}$

Let $\displaystyle s = 0 \implies 11\,B = 3 \implies B = \frac{3}{11}$

Let $\displaystyle s = -11 \implies 121\,C = 3 \implies C = \frac{3}{121}$

Then $\displaystyle A\,s\left( s + 11 \right) + \frac{3}{11} \left( s + 11 \right) + \frac{3}{121}\,s^2 \equiv 3$

Let $\displaystyle s = 1$

$\displaystyle \begin{align*}

12\,A + \frac{36}{11} + \frac{3}{121} &= 3 \\

12\,A + \frac{396}{121} + \frac{3}{121} &= \frac{33}{121} \\

12\,A &= -\frac{366}{121} \\

A &= -\frac{61}{242}

\end{align*}$

So

$\displaystyle \begin{align*}

Y\left( s \right) &= -\frac{61}{242}\left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{3}{121} \left( \frac{1}{s + 11} \right) + 5 \left( \frac{1}{s + 11 } \right) \\

Y\left( s \right) &= -\frac{61}{242} \left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{608}{121} \left( \frac{1}{s + 11} \right) \\

\\

y\left( t \right) &= -\frac{61}{242} + \frac{3}{11}\,t + \frac{608}{121}\,\mathrm{e}^{-11\,t}

\end{align*}$