Applications of Laplace Transform to Ordinary Differential Equations

1. Solve the differential equation: \[ y''(t) + 4y'(t) + 3y = e^{-t}, \quad y(0) = y'(0) = 1 \] using Laplace transforms.

Solution: Given the differential equation: \[ y''(t) + 4y'(t) + 3y = e^{-t}, \quad y(0) = y'(0) = 1 \]

Taking the Laplace transform on both sides, we get: \[ L\{y''(t) + 4y'(t) + 3y\} = L\{e^{-t}\} \] \[ \Rightarrow L\{y''(t)\} + 4L\{y'(t)\} + 3L\{y\} = \frac{1}{s+1} \] Substituting the Laplace transforms of derivatives: \[ \left[ s^2 L\{y(t)\} - s y(0) - y'(0) \right] + 4 \left[ s L\{y(t)\} - y(0) \right] + 3 L\{y(t)\} = \frac{1}{s+1} \] Using the initial conditions \( y(0) = 1 \) and \( y'(0) = 1 \): \[ \left[ s^2 L\{y(t)\} - s - 1 \right] + 4 \left[ s L\{y(t)\} - 1 \right] + 3 L\{y(t)\} = \frac{1}{s+1} \] Simplifying: \[ L\{y(t)\} (s^2 + 4s + 3) = \frac{1}{s+1} + s + 5 \] \[ L\{y(t)\} = \frac{s^2 + 6s + 6}{(s+1)(s^2 + 4s + 3)} = \frac{s^2 + 6s + 6}{(s+1)^2 (s+3)} \]

Resolve into partial fractions: \[ \frac{s^2 + 6s + 6}{(s+1)^2 (s+3)} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{s+3} \] Multiply through by the denominator: \[ s^2 + 6s + 6 = A(s+1)(s+3) + B(s+3) + C(s+1)^2 \] Solve for \( A \), \( B \), and \( C \):

• Set \( s = -1 \): \( B = \frac{1}{2} \)
• Set \( s = -3 \): \( C = -\frac{3}{4} \)
• Compare coefficients of \( s^2 \): \( A + C = 1 \Rightarrow A = \frac{7}{4} \)

Thus: \[ L\{y(t)\} = \frac{7}{4(s+1)} + \frac{1}{2(s+1)^2} - \frac{3}{4(s+3)} \] Taking the inverse Laplace transform: \[ y(t) = \frac{7}{4} e^{-t} + \frac{1}{2} t e^{-t} - \frac{3}{4} e^{-3t} \]

Therefore, the solution is: \[ y(t) = \frac{7}{4} e^{-t} + \frac{1}{2} t e^{-t} - \frac{3}{4} e^{-3t} \]

2. Solve the differential equation: \[ y'' + 2y' + 5y = e^{-t} \sin t, \quad y(0) = 0, \, y'(0) = 1 \] using Laplace transforms.

Solution: Given the differential equation: \[ y'' + 2y' + 5y = e^{-t} \sin t, \quad y(0) = 0, \, y'(0) = 1 \]

Taking the Laplace transform on both sides, we get: \[ L\{y'' + 2y' + 5y\} = L\{e^{-t} \sin t\} \] \[ \Rightarrow L\{y''(t)\} + 2L\{y'(t)\} + 5L\{y\} = L\{e^{-t} \sin t\} \] Substituting the Laplace transforms of derivatives: \[ \left[ s^2 L\{y(t)\} - s y(0) - y'(0) \right] + 2 \left[ s L\{y(t)\} - y(0) \right] + 5 L\{y(t)\} = L\{\sin t\}_{s \to (s+1)} \] Using the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \): \[ \left[ s^2 L\{y(t)\} - 1 \right] + 2s L\{y(t)\} + 5 L\{y(t)\} = \frac{1}{(s+1)^2 + 1} \] Simplifying: \[ L\{y(t)\} (s^2 + 2s + 5) = \frac{1}{(s+1)^2 + 1} + 1 = \frac{s^2 + 2s + 3}{s^2 + 2s + 2} \] \[ L\{y(t)\} = \frac{s^2 + 2s + 3}{(s^2 + 2s + 2)(s^2 + 2s + 5)} \]

Rewrite the numerator and denominator: \[ \frac{s^2 + 2s + 3}{(s^2 + 2s + 2)(s^2 + 2s + 5)} = \frac{(s+1)^2 + 2}{[(s+1)^2 + 1][(s+1)^2 + 4]} \] Let \( p = (s+1)^2 \), then: \[ \frac{p + 2}{(p + 1)(p + 4)} = \frac{A}{p + 1} + \frac{B}{p + 4} \] Resolving into partial fractions: \[ \frac{p + 2}{(p + 1)(p + 4)} = \frac{1}{3(p + 1)} + \frac{2}{3(p + 4)} \] Substituting back \( p = (s+1)^2 \): \[ \frac{(s+1)^2 + 2}{[(s+1)^2 + 1][(s+1)^2 + 4]} = \frac{1}{3[(s+1)^2 + 1]} + \frac{2}{3[(s+1)^2 + 4]} \]

Taking the inverse Laplace transform: \[ y(t) = e^{-t} L^{-1}\left\{ \frac{s^2 + 2}{(s^2 + 1)(s^2 + 4)} \right\} \] \[ = e^{-t} \left[ \frac{1}{3} L^{-1}\left\{ \frac{1}{s^2 + 1} \right\} + \frac{2}{3} L^{-1}\left\{ \frac{1}{s^2 + 4} \right\} \right] \] \[ = e^{-t} \left[ \frac{1}{3} \sin t + \frac{2}{3} \cdot \frac{1}{2} \sin 2t \right] \] \[ = \frac{e^{-t}}{3} \left[ \sin t + 2 \sin 2t \right] \]

Therefore, the solution is: \[ y(t) = \frac{e^{-t}}{3} \left[ \sin t + 2 \sin 2t \right] \]

Model Problems

Solve the following differential equations using Laplace transforms:

1. \[ y''' + 2y'' - y' - 2y = 0, \quad y(0) = y'(0) = 0, \quad y''(0) = 6 \]
2. \[ y'' - 2y' - 8y = 0, \quad y(0) = 3, \quad y'(0) = 6 \]
3. \[ y'' + y = 2e^t, \quad y(0) = 0, \quad y'(0) = 2 \]
4. \[ \frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 3y = \sin t, \quad y = \frac{dy}{dt} = 0 \text{ at } t = 0 \]
5. \[ y''' - 3y'' + 3y' - y = t^2 e^t, \quad y(0) = 1, \quad y'(0) = 0, \quad y''(0) = -2 \]
6. \[ (D^2 + n^2)x = a \sin(nt + \alpha), \quad x = Dx = 0 \text{ at } t = 0 \]
7. \[ (D^2 + 9)x = \cos 2t, \quad x(0) = 1, \quad x\left(\frac{\pi}{2}\right) = -1 \]
8. \[ \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + x = e^t, \quad x = 2, \frac{dx}{dt} = -1 \text{ at } t = 0 \]
9. \[ \frac{d^2x}{dt^2} + 9x = \sin t, \quad x(0) = 1, \quad x\left(\frac{\pi}{2}\right) = 1 \]
10. \[ (D^2 + 1)x = t \cos 2t, \quad x = 0, \frac{dx}{dt} = 0 \text{ at } t = 0 \]
11. \[ (D^2 + 4D + 5)y = 5, \quad y(0) = y'(0) = 0 \]
12. \[ \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + x = 3t e^{-t}, \quad x(0) = 4, \quad x'(0) = 0 \]
13. \[ x'' + 4x' = 9t, \quad x(0) = 0, \quad x'(0) = 7 \]
14. \[ y''' + 4y'' + 5y' + 2y = 10 \cos t, \quad y(0) = y'(0) = 0, \quad y''(0) = 3 \]

Answers

1. \[ y(t) = e^t - 3e^{-t} + 2e^{-2t} \]
2. \[ y(t) = 2e^{4t} + e^{-2t} \]
3. \[ y(t) = e^t + \cos t + \sin t \]
4. \[ y(t) = \frac{1}{8}e^t - \frac{1}{40}e^{-3t} - \frac{1}{10}\cos t - \frac{1}{5}\sin t \]
5. \[ y(t) = \frac{e^t}{60} \left( t^5 - 30t^2 - 60t + 60 \right) \]
6. \[ x(t) = \frac{a}{2n^2} \left[ \cos \alpha \sin nt - nt \cos(nt + \alpha) \right] \]
7. \[ x(t) = \frac{1}{5} \left[ \cos 2t + 4 \cos 3t + 4 \sin 3t \right] \]
8. \[ x(t) = 2e^t - 3t e^t + \frac{t^2}{2} e^t \]
9. \[ x(t) = \frac{1}{24} \left[ 3 \sin t - 21 \sin 3t + 24 \cos 3t \right] \]
10. \[ x(t) = \frac{1}{9} \left[ 4 \sin 2t - 3t \cos 2t - 5 \sin t \right] \]
11. \[ x(t) = 1 - e^{-2t} (\cos t + 2 \sin t) \]
12. \[ x(t) = \frac{e^{-t}}{2} (t^3 + 8t + 8) \]
13. \[ x(t) = \frac{1}{8} (18t + 19 \sin 2t) \]
14. \[ y(t) = 2e^{-t} - e^{-2t} - 2t e^{-t} - \cos t + 2 \sin t \]