Use the Laplace transform?

Question

Use the Laplace transform to solve y^(4)-4y'''+6y"-4y'+y=0; y(0)=0, y'(0)=1, y"(0)=0, y'''(0)=1.

Answer: y=te^t-t^2*e^t+(2/3)t^3*e^t

L(y^(4))-4L(y''')+6L(y")-4L(y')+L(y)=L(0)

But I don't know what's L(y^(4)) and L(y''').

Answer 1

This si the amswer to your last question about (s² -4s +7)/(s-1)⁴. Let's rewrite the numerator:

 s² -4s +7 = s² - 2s +1 -1 -2s +7 = (s-1)² -2(s-1) + 4                (1)

If we divide (1) by (s-1) ⁴ we will  obtain:

            1/(s-1)²  - 2 /(s-1) ³  + 4/(s-1)⁴                                               (2)

But 1/(s-1) ⁿ  is a Laplace trabnform of the function [1/(n-1)! t ^n-1 e^t.

If you consequitively apply this formula for n = 2, 3, 4 you will come up with function

                                         y(t) = te^t - t² e^t +(2/3) t³ e^t