Use the Laplace transform to solve the given initial-value problem. y''-10y'+25y=t, y(0)=0, y'(0)=1 y(t)=?

Apply L to both sides:
[s^2 Y(s) - 0s - 1] - 10 [s Y(s) - 0] + 25 Y(s) = 1/s^2.

Solve for Y(s):
(s^2 - 10s + 25) Y(s) = 1 + 1/s^2
==> (s - 5)^2 Y(s) = 1 + 1/s^2
==> Y(s) = 1/(s - 5)^2 + 1/(s^2 (s - 5)^2).

By partial fractions, we can rewrite this as
Y(s) = 1/(s - 5)^2 + (1/125) [5/s^2 + 2/s - 2/(s - 5) + 5/(s - 5)^2]
........= (1/125) [5/s^2 + 2/s - 2/(s - 5) + 130/(s - 5)^2].

Inverting yields
y(t) = (1/125) [5t + 2 - 2e^(5t) + 130te^(5t)].

I hope this helps!