The given initial-value problem.y″ − 2y′ − 8y=5y(0)=1y′(0)=0

Question

The given initial-value problem.y″ − 2y′ − 8y=5y(0)=1y′(0)=0

Margot Mill · Accepted Answer

To solve, we use the properties of the Laplace transform and the Laplace table
The Laplace transform satisfies the linearity properties

L [f = g] = L [f] + L [g]

L [c f] = c L [f]

For all transfotmable functions f and g and constants c.
Laplace transform:

L [y^{″} - 2 y^{'} - 8 y] = L [5]

L [y^{″}] - 2 L [y^{'}] - 8 L [y] = 5 L [1]

Undefined control sequence \8

Y (s) (s^{2} - 2 s - 8) - y^{'} (0) + y (0) (2 - s) = \frac{5}{s}

Substitute initial conditions,

y (0) = 1

y^{'} (0) = 0

Solve for

Y (s)

Y (s) (s^{2} - 2 s - 8) - 0 + 1 (2 - s) = \frac{5}{s}

Y (s) (s^{2} - 2 s - 8) - s + 2 = \frac{5}{s}

Y (s) (s^{2} - 2 s - 8) = \frac{5}{s} - (- s + 2)

= \frac{5 - (- s + 2) s}{s}

Y (s) = \frac{s^{2} - 2 s + 5}{s (s^{2} - 2 s - 8)}

= \frac{5}{8 s} + \frac{13}{12 (s + 2)} + \frac{13}{24 (s - 4)}

Inverse transform:

L^{- 1} [Y (s)] = L^{- 1} [- \frac{5}{8 s} + \frac{13}{12 (s + 2)} + \frac{13}{24 (s - 4)}]

= - \frac{5}{8} L^{- 1} [\frac{1}{s}] + \frac{13}{12} L^{- 1} [\frac{1}{s - (- 2)}] + \frac{13}{24} L^{-} [\frac{1}{s - 4}]

y (t) = - \frac{5}{8} + \frac{13}{12} e^{- 2 t} + \frac{13}{24} e^{4 t}

The given initial-value problem. y'' - 2y' - 8y = 5 y(0) = 1 y' (0) = 0