Answer:
a) y₁(t) = (2.5e⁰•⁸ᵗ - 2.5e⁰•⁴ᵗ)
b) y₂(t) = (2e⁰•⁴ᵗ - e⁰•⁸ᵗ)
c) W(y₁,y₂) = -e¹•²ᵗ
Step-by-step explanation:
To solve, 25y" - 30y' + 8y = 0
With conditions, y₁'(0) = 1 and y₁(0) = 0 for y₁
And y₂'(0) = 0 and y₂(0) = 1 for y₂
Solving for y₁ first, using Laplace transforms.
Note:
L(y") = s² Y(s) - sy(0) - y'(0)
L(y') = s Y(s) - y(0)
L(y₁") = s² Y(s) - sy₁(0) - y'₁(0) = s² Y(s) - 0 - 1 = s² Y(s) - 1
L(y₁') = s Y(s) - y₁(0) = s Y(s)
L [25y" - 30y' + 8y] = L (0)
25(s²Y(s) - 1) - 30(sY(s)) + 8Y(s) = 0
25s²Y(s) - 25 - 30sY(s) + 8Y(s) = 0
Y(s) [25s² - 30s + 8] - 25 = 0
Y(s) [25s² - 30s + 8] = 25
Y(s) = (25)/[25s² - 30s + 8]
Y(s) = (25)/[(5s-4)(5s-2)]
To continue, we need to resolve 1/[(5s-4)(5s-2)] into partial fractions.
The calculation is continued on the attached images.
