Answer to Question #117545 in Complex Analysis for MOULYA T

Question

find the laplace transform of y"-4y'+9y=t, y(0)=0,y'(0)=1

Answer 1

The given equation is y'' - 4y' + 9y = t, and y(0) = 0 and y'(0) = 1.

We know, L[t] = 1/s²

And, L[(dⁿ/dtⁿ)y(t)] = sⁿ L[y(t)] - s^n-1 y(0) - s^n-2 y'(0) - ..... - y^n-1(0)

Calculating L[y"] :

L[y"] = s² L[y] - s¹ y(0) - s⁰ y'(0)

i.e., L[y"] = s² L[y] - 0 - 1

L[y"] = s² L[y] - 1

Calculating L[y'] :

L[y'] = s¹ L[y] - s⁰ y(0)

L[y'] = s L[y]

Rewriting the given equation as,

y" - 4y' + 9y - t = 0

Calculating the laplace transform :

L[ y" - 4y' + 9y - t] = L[y"] - 4 L[y'] + 9 L[y] - L[t] = 0

(s² L[y] - 1) - 4(sL[y]) + 9 L[y] - (1/s²) = 0

(s² - 4s + 9)L[y] - 1 - 1/s² = 0

(s² - 4s + 9)L[y] = 1 + 1/s²

(s² - 4s + 9)L[y] = (s² + 1)/s²

L[y] = (s² + 1)/(s²(s² - 4s + 9))

$\therefore$ L[y] = 1/(s² - 4s + 9) + 1/(s²(s² - 4s + 9))