Question

laplace transform to solve the following ode;

4y''+y'-3y=2e^(-5t)

initial conditions y'(0)=0 y(0)=1

Answer 1

Answer on Question #82482 – Math – Differential Equations

Question

Laplace transform to solve the following ode

$4y'' + y' - 3y = 2e^{-5t}$ initial conditions $y'(0) = y(0) = 1$ .

Solution

Let Y(s) be the Laplace transform of y(t). Taking the Laplace transform of the differential equation we have:

$$L[4y'' + y' - 3y] = L[2e^{-5t}].$$

Now

$$\begin{array}{l} L[4y'' + y' - 3y] = 4L[y''] + L[y'] - 3L[y] = 4 \cdot (s^2 \cdot Y(s) - s \cdot y(0) - y'(0)) + \\ + s \cdot Y(s) - y(0) - 3 \cdot Y(s) = 4 \cdot s^2 \cdot Y(s) - 4 \cdot s \cdot 1 - 4 \cdot 1 + s \cdot Y(s) - 1 - 3 \cdot Y(s) = \\ = Y(s) \cdot (4s^2 + s - 3) - 4s - 5; \\ \end{array}$$ $$L[2e^{-5t}] = 2 \cdot \frac{1}{s + 5} = \frac{2}{s + 5}.$$

Then

$$Y(s) \cdot (4s^2 + s - 3) - 4s - 5 = \frac{2}{s + 5};$$ $$Y(s) \cdot (4s^2 + s - 3) = \frac{2}{s + 5} + 4s + 5;$$ $$Y(s) \cdot (4s^2 + s - 3) = \frac{2 + (4s + 5) \cdot (s + 5)}{s + 5};$$ $$Y(s) = \frac{2 + 4s^2 + 20s + 5s + 25}{(s + 5) \cdot (4s^2 + s - 3)};$$ $$Y(s) = \frac{4s^2 + 25s + 27}{(s + 5) \cdot (4s^2 + s - 3)};$$ $$Y(s) = \frac{4s^2 + 25s + 27}{(s + 5) \cdot (s + 1) \cdot (4s - 3)}.$$

We can simplify this expression using the method of partial fractions.

$$\begin{array}{l} \frac{4s^2 + 25s + 27}{(s + 5) \cdot (s + 1) \cdot (4s - 3)} = \frac{A}{s + 5} + \frac{B}{s + 1} + \frac{C}{4s - 3} = \\ = \frac{A \cdot (s + 1) \cdot (4s - 3) + B \cdot (s + 5) \cdot (4s - 3) + C \cdot (s + 1) \cdot (s + 5)}{(s + 5) \cdot (s + 1) \cdot (4s - 3)} = \\ = \frac{A \cdot \left(4s^2 - 3s + 4s - 3\right) + B \cdot \left(4s^2 - 3s + 20s - 15\right) + C \cdot \left(s^2 + 5s + s + 5\right)}{(s + 5) \cdot (s + 1) \cdot (4s - 3)} = \\ = \frac{A \cdot (4s^2 + s - 3) + B \cdot (4s^2 + 17s - 15) + C \cdot (s^2 + 6s + 5)}{(s + 5) \cdot (s + 1) \cdot (4s - 3)} = \\ = \frac{s^2 \cdot (4A + 4B + C) + s \cdot (A + 17B + 6C) + (-3A - 15B + 5C)}{(s + 5) \cdot (s + 1) \cdot (4s - 3)}; \end{array}$$ $$s^2: \quad 4A + 4B + C = 4,$$ $$s^1: \quad A + 17B + 6C = 25,$$ $$s^0: \quad -3A - 15B + 5C = 27.$$

We have a system:

\begin{array}{l} \left\{ \begin{array}{l} 4A + 4B + C = 4; \\ A + 17B + 6C = 25; \\ -3A - 15B + 5C = 27; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C = 4 - 4A - 4B; \\ A + 17B + 6 \cdot (4 - 4A - 4B) = 25; \\ -3A - 15B + 5C = 27; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C = 4 - 4A - 4B; \\ A + 17B + 24 - 24A - 24B = 25; \\ -3A - 15B + 5C = 27; \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} C = 4 - 4A - 4B; \\ -23A - 7B = 1; \\ -3A - 15B + 5C = 27; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C = 4 - 4A - 4 \cdot \left(-\frac{23}{7} A - \frac{1}{7}\right); \\ B = -\frac{23}{7} A - \frac{1}{7}; \\ -3A - 15B + 5C = 27; \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} C = \frac{64}{7} A + \frac{32}{7}; \\ B = -\frac{23}{7} A - \frac{1}{7}; \\ -3A - 15 \cdot \left(-\frac{23}{7} A - \frac{1}{7}\right) + 5 \cdot \left(\frac{64}{7} A + \frac{32}{7}\right) = 27; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C = \frac{64}{7} A + \frac{32}{7}; \\ B = -\frac{23}{7} A - \frac{1}{7}; \\ -3A + \frac{345}{7} A + \frac{15}{7} + \frac{320}{7} A + \frac{160}{7} = 27; \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} C = \frac{64}{7} A + \frac{32}{7}; \\ B = -\frac{23}{7} A - \frac{1}{7}; \\ \frac{644}{7} A = \frac{14}{7}; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C = \frac{768}{161}; \\ B = -\frac{3}{14}; \\ A = \frac{1}{46}. \end{array} \right. $$\begin{array}{l} \frac{4s^{2} + 25s + 27}{(s + 5) \cdot (s + 1) \cdot (4s - 3)} = \frac{A}{s + 5} + \frac{B}{s + 1} + \frac{C}{4s - 3} = \frac{1}{46} \cdot \frac{1}{s + 5} - \frac{3}{14} \cdot \frac{1}{s + 1} + \\ + \frac{768}{161} \cdot \frac{1}{4s - 3} = \frac{1}{46} \cdot \frac{1}{s + 5} - \frac{3}{14} \cdot \frac{1}{s + 1} + \frac{768}{161 \cdot 4} \cdot \frac{1}{s - \frac{3}{4}} = \frac{1}{46} \cdot \frac{1}{s + 5} - \\ - \frac{3}{14} \cdot \frac{1}{s + 1} + \frac{192}{161} \cdot \frac{1}{s - \frac{3}{4}}. \end{array}$$ $$Y(s) = \frac{1}{46} \cdot \frac{1}{s + 5} - \frac{3}{14} \cdot \frac{1}{s + 1} + \frac{192}{161} \cdot \frac{1}{s - \frac{3}{4}};$$ $$\begin{array}{l} \frac{1}{s + 5} \rightarrow e^{-5t}, \\ \frac{1}{s + 1} \rightarrow e^{-t}, \\ \frac{1}{s - \frac{3}{4}} \rightarrow e^{\frac{3}{4}t}. \\ Y(s) \rightarrow y(t). \end{array}$$

One finally gets

$$y(t) = \frac{1}{46} \cdot e^{-5t} - \frac{3}{14} \cdot e^{-t} + \frac{192}{161} \cdot e^{\frac{3}{4}t}.$$

Answer: $y(t) = \frac{1}{46} \cdot e^{-5t} - \frac{3}{14} \cdot e^{-t} + \frac{192}{161} \cdot e^{\frac{3}{4}t}$.

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