Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Solve the initial value problems using Laplace transforms   y'' + y' - 2y = 4e^2t            y(0) = 1;    y'(0) = -1?

y'' + y' - 2y = 4e^2t          y(0) = 1;     y'(0) = -1

Relevance
• 1 month ago

Given: y(0) = 1, y'(0) = -1

Second Order Linear Ordinary Differential Equation:

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

ℒ(y") + ℒ(y') - ℒ(2y) = ℒ(4e^2t)

Use Laplace transform rules found in the cited source:

s²ℒ(y) - sy(0) - y'(0) + sℒ(y) - y(0) - 2ℒ(y) = 4ℒ(e^2t)

s²ℒ(y) - s(1) - (-1) + sℒ(y) - (1) - 2ℒ(y) = 4ℒ(e^2t)

ℒ(y)[s² + s - 2] - s = 4 / (s - 2)

ℒ(y) = [4 / (s - 2)(s² + s - 2)] + [s / (s² + s - 2)]

ℒ(y) = (4 + s(s - 2) / (s - 2)(s² + s - 2))

ℒ(y) = ((4 + s² - 2s) / (s - 2)(s² + s - 2))

y = ℒ⁻¹((s² - 2s + 4) / (s - 2)(s - 1)(s + 2))

y = ℒ⁻¹[A / (s - 2)] + ℒ⁻¹[B / (s - 1)] + ℒ⁻¹[C / (s + 2)]

Partial Fraction Expansion:

(s² - 2s + 4) / (s - 2)(s - 1)(s + 2) = [A / (s - 2)] + [B / (s - 1)] + [C / (s + 2)]

A(s - 1)(s + 2) + B(s - 2)(s + 2) + C(s - 2)(s - 1) = (s² - 2s + 4)

A(s² + s - 2) + B(s² - 4) + C(s² - 3s + 2) = (s² - 2s + 4)

A + B + C = 1

A - 3C = -2

-2A - 4B + 2C = 4

3 Equations, 3 Unknowns:

A = -2 + 3C

B = 1 - C - A

B = 1 - C - (-2 + 3C)

B = 3 - 4C

-2(-2 + 3C) - 4(3 - 4C) + 2C = 4

4 - 6C - 12 + 16C + 2C = 4

12C = 12

C = 1

A = -2 + 3(1) = 1

B = 3 - 4(1) = -1

Plug back in to the inverse Laplace equation:

y = ℒ⁻¹[(1) / (s - 2)] + ℒ⁻¹[(-1) / (s - 1)] + ℒ⁻¹[(1) / (s + 2)]

Use inverse Laplace transform rules found in the cited source:

y = e^(2t) - e^t + e^(-2t)

• 1 month ago

y(t) = ⅓ e⁽⁻²ᵗ⁾ (-3 e⁽²ᵗ⁺²⁾ (2 t + 1) + e³ᵗ + 4 e⁽³ᵗ⁺²⁾  - e² + 2)

• rotchm
Lv 7
1 month ago

Apply the exact same procedure already given in: