Question

Using the method of Laplace transform, solve the following initial value problem:
y′′-3y′+2y=4e²ᵗ ;
y(0)=-3 ;
y′(0)=5 ;

Answer 1

Answer on Question #84503 – Math – Differential Equations

Question

**Task:** Using the method of Laplace transform, solve the following initial value problem:

$$\begin{array}{l} y^{\prime \prime} - 3y^{\prime} + 2y = 4e^{2t} \\ y(0) = -3 \\ y^{\prime}(0) = 5 \\ \end{array}$$

Solution

$$s^{2} \cdot Y(s) - s \cdot y(0) - y^{\prime}(0) - 3 \cdot \left(s \cdot Y(s) - y(0)\right) + 2 \cdot Y(s) = \frac{4}{s - 2}$$ $$Y(s) = \frac{(14 - s)(s - 2) + 4}{(s - 2)^{2}(s - 1)} = \frac{-3s^{2} + 20s - 24}{(s - 2)^{2}(s - 1)} = \frac{A}{s - 2} + \frac{B}{(s - 2)^{2}} + \frac{C}{s - 1} \Rightarrow$$ $$\left\{ \begin{array}{l} A + C = -3 \\ B - 3A - 4C = 20 \\ 2A - B + 4C = -24 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} A = 4 \\ B = 4 \\ C = -7 \end{array} \right. \Rightarrow$$ $$Y(s) = \frac{4}{s - 2} + \frac{4}{(s - 2)^{2}} - \frac{7}{s - 1}$$

Using inverse Laplace transform:

$$y(t) = e^{2t} \cdot (4 + 4t) - 7e^{t}$$

**Answer:** $y(t) = e^{2t} \cdot (4 + 4t) - 7e^{t}$.

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