Question

Find the function y1 of t which is the solution of

Answer 1

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.