Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the convolution theorem to obtain a formula for the solution to the given initial value problem, where $\mathit{g}\left(t\right)$is piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ and of exponential order.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{5}\mathit{y}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{;}\mathbf{ }\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}$

Step-by-Step Solution

Answer

The solution to the given initial value problem by using a convolution theorem to obtain a formula is.

 $y\left(t\right) ={\int }_0^t{e}^{-2\left(t-v\right)}\sin (t - v)g\left(v\right)dv + e^{- 2t}\cos t + 3e^{- 2t}\sin t$

Step 1: Define convolution theorem

 

The convolution theorem states that under suitable conditions the Fourier transform of a convolution of two functions is the point wise product of their Fourier transforms.

 

Step 2: Use Laplace transform and simplify the given equation:

Consider the equation,

$y\text{'}\text{'} + 4y\text{'} + 5y = g\left(t\right)$

Apply Laplace transform and its linearity in the equation,

$$\begin{gathered} L \{y\text{'}\text{'} + 4y\text{'} + 5y\}\left(s\right) = L\left\{g\left(t\right)\right\}\left(s\right) \\ L \left\{y\right\}\left(s\right) + 4\left\{L\right\}\{y\text{'}\}\left(s\right) + 5\left\{L\right\}\{ y\} \left(s\right) = G\left(s\right) \\ {s}^{2}Y\left(s\right) - sy\left(0\right) - y\text{'}\left(0\right) + 4\left[sY\right(s) - y(0\left)\right] + 5Y\left(s\right) = G \\ {s}^{2}Y\left(s\right) - s - 1 + 4sY\left(s\right) - 4 + 5Y\left(s\right) = G\left(s\right) \end{gathered}$$

$$\begin{gathered} (s^2 + 4s + 5)Y\left(s\right) = G\left(s\right) + s + 5 \\ Y\left(s\right)= \frac{G\left(s\right)}{s^2+4s+5}+\frac{s + 5}{s^2 + 4s + 5} \\ Y\left(s\right)= \frac{G\left(s\right)}{{\left(s+2\right)}^2+1}+\frac{s + 2}{{\left(s+2\right)}^2+1}+\frac{ 3}{{\left(s+2\right)}^2+1} \end{gathered}$$

Apply inverse Laplace transform, 

$$\begin{gathered} y\left(t\right) =L^{-1}\left\{ \frac{G\left(s\right)}{{\left(s+2\right)}^2+1}+\frac{s + 2}{{\left(s+2\right)}^2+1}+\frac{ 3}{{\left(s+2\right)}^2+1}\right\} \\ =L^{-1} \left\{\frac{G\left(s\right)}{{\left(s+2\right)}^2+1}\right\}+L^{-1}\left\{\frac{s + 2}{{\left(s+2\right)}^2+1}\right\}+L^{-1}\left\{\frac{ 3}{{\left(s+2\right)}^2+1}\right\} \\ =L^{-1} \left\{\frac{G\left(s\right)}{{\left(s+2\right)}^2+1}\right\}+e^{-2t}\cos t+3e^{-2t}\sin t \end{gathered}$$

Thus, we get

$L^{-1} \left\{\frac{1}{{\left(s+2\right)}^2+1}\right\}=e^{-2t}\sin t$

 

Step 3: Simplify the equation using the convolution formula

Hence, we have

$L^{-1} \left\{\frac{1}{{\left(s+2\right)}^2+1}\left\{G\right(s)\right\}= e^{2t}\sin t * g\left(t\right)$

Use convolution formula,

$(f * g)\left(t\right)={\int }_0^t\left\{f\right(t - v\left)g\right(v)dv$

$L^{-1}\left\{\frac{1}{{(s + 2)}^2 + 1}G\right\}={\int }_{}^{}{e}^{-2\left(t-v\right)}\sin (t - v)g\left(v\right)dv$

Therefore, by using this formula, the equation can be written as

$$\begin{gathered} y\left(t\right)=L^{-1}\left\{\frac{G\left(s\right)}{{\left(s+2\right)}^2+1}\right\}+e^{-2t}c\mathrm{os} t + 3e^{-2t}\sin t \\ =L^{-1}\left\{\frac{1}{{\left(s+2\right)}^2+1}G\left(s\right)\right\}+e^{-2t}c\mathrm{os} t + 3e^{-2t}\sin t \\ ={\int }_0^t{e}^{-2\left(t-v\right)}s\mathrm{in} (t - v)g\left(v\right)dv +dv+e^{-2t}c\mathrm{os} t + 3e^{-2t}\sin t \end{gathered}$$

Thus, the required solution to obtain a formula  for the given equation is. 

$y\left(t\right)={\int }_0^t{e}^{-2\left(t-v\right)}s\mathrm{in} (t - v)g\left(v\right)dv +dv+e^{-2t}c\mathrm{os} t + 3e^{-2t}\sin t$