Question

1. Use the Laplace transform to solve the given initial value problem.

y^4) − 81y = 0; y(0) = 20, y′(0) = 39, y′′(0) = 108, y′′′ (0) =1 89


Enclose arguments of functions in parentheses. For example, sin(2x).

y(t) = ___________


2. Find the Laplace transform of the function:

f(t) = { 0, t < 4
{ t^2 - 8t + 9, t is greater than or equal 4

Answer 1

1) Given differential equation is $y^4 − 81y = 0 : y(0) = 20, y'(0) = 39, y''(0) = 108, y'''(0) =1 89$ .

Taking Laplace Transform, we get

$L(y^4-81y ) = 0$

Now using Linear property of Laplace, we get

$L(y^4) - 81 L(y) = 0$ .

$\implies (s^4 Y-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0)) - 81 Y = 0$ where $L(y(t)) = Y(s)$ .

$\implies (s^4 Y-20s^3-39s^2-108s-189) - 81 Y = 0$

$\implies (s^4-81)Y = 20s^3+39s^2+108s+189$

$\implies Y = \frac{20s^3+39s^2+108s+189}{s^4-81}$ .

Now, taking Laplace Inverse transformation on both sides, we get

$y = L^{-1} (\frac{20s^3+39s^2+108s+189}{s^4-81})$

$=L^{-1}\left\{\frac{4s+9}{s^2+9}+\frac{3}{s+3}+\frac{13}{s-3}\right\} \\ =L^{-1}\left\{\frac{4s}{s^2+9}+\frac{9}{s^2+9}+\frac{3}{s+3}+\frac{13}{s-3}\right\} \\ =4L^{-1}\left\{\frac{s}{s^2+9}\right\}+9L^{-1}\left\{\frac{1}{s^2+9}\right\}+3L^{-1}\left\{\frac{1}{s+3}\right\}+13L^{-1}\left\{\frac{1}{s-3}\right\} \\ =4\cos \left(3t\right)+9\cdot \frac{1}{3}\sin \left(3t\right)+3e^{-3t}+13e^{3t}$

So, $y =4\cos \left(3t\right)+3\sin \left(3t\right)+3e^{-3t}+13e^{3t}$ .

2) $L \{f(t)H(t− a)\}=e^{−as}L \{f(t+ a)\}$

Hence, $L\{ (t^2-8t+9 ) H(t-4) \} = e^{-4s} L\{ (t+4)^2-8(t+2)+9 \}$

$= e^{-4s} L\{ t^2 +9 \} = e^{-4s} (\frac{2}{s^3}+\frac{9}{s})$ .