Q1E

Question

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

\(y'' - 2y' + y = g(t);\quad y(0) = - 1,\quad y'(0) = 1\)

Step-by-Step Solution

Verified

Answer

The solution to the given initial value problem by using convolution theorem is

1Step 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.

2Step 2: Use Laplace and inverse Laplace transform to the given values

Given

\(y'' - 2y' + y = g(t),\quad y(0) = - 1,\quad y'(0) = 1\)

Applying Laplace transform on both sides we get

\(\mathcal{L}\left[ {y'' - 2y' + y} \right] = \mathcal{L}[g(t)]\)

\(\begin{array}{c}{s^2}y(s) - sy(0) - y'(0) - 2[sy(s) - y(0)] + y(s) = g(s)\\{s^2}y(s) + s - 1 - 2[sy(s) + 1] + y(s) = g(s)\\\left( {{s^2} - 2s + 1} \right)y(s) = g(s) - (s - 3)\\y(s) = \frac{{g(s)}}{{{s^2} - 2s + 1}} - \frac{{s - 3}}{{{s^2} - 2s + 1}}\end{array}\)

Using partial fraction we get

\(\frac{{s + 3}}{{{s^2} - 2s + 1}} = \frac{1}{{s - 1}} - \frac{2}{{{{(s - 1)}^2}}}\)

Equation first becomes as

\(y(s) = \frac{{g(s)}}{{{{(s - 1)}^2}}} - \frac{1}{{s - 1}} + \frac{2}{{{{(s - 1)}^2}}}\)

Taking inverse Laplace transform on both sides we get,

\({\mathcal{L}^{ - 1}}[y(s)] = {\mathcal{L}^{ - 1}}\left[ {\frac{{g(s)}}{{{{(s - 1)}^2}}}} \right] - {\mathcal{L}^{ - 1}}\left[ {\frac{1}{{s - 1}}} \right] + {\mathcal{L}^{ - 1}}\left[ {\frac{2}{{{{(s - 1)}^2}}}} \right] \cdots (2)\)

Using formula

Where

\(\begin{array}{c}f(s) = \frac{1}{{{{(s - 1)}^2}}}\\f(t) = t{e^t}\end{array}\)

Equation (2) becomes

Q20E

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