use the laplacr transform to solve the following initial val

use the laplacr transform to solve the following initial value problems for second order equations.

show complete answer please.

(f) y\" + 4y\' + 3y = cosh 2t, y(0) = 0 = y\'(0).

Solution

Given equation : y\" + 4y\' + 3y = cosh 2t , y\'(0) = 0 = y(0)

Taking Laplace transform on both sides

L{ y\" + 4y\' + 3y } = L {cosh 2t}

L{y\" } + 4 L {y\' } + 3 L{y} = L { cosh 2t}

s^2 L{y} - s y(0) - y\' (0) + 4 ( s L{y} - y(0) ) + 3 L{y} = L{ cosh 2t }

(s^2 + 4s + 3 ) L{y} = L { cosh 2t}

we know laplace transform of L(cosh (at)) = s / ( s^2 - a^2 )

( s^2 + 4s + 3 ) L{y} = s / s^2 - 4

L{y} = s / ( s^2 - 4) ( s^2 + 4s +3)

Expanding this last term in partial fractions gives

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

= A / (s + 2 ) + B/ (s-2 ) + C / (s +3) + D / (s + 1)

plugging in s = -2 gives A = -1 / 2

s = 2 gives B = 1/ 30

s = -3 gives C = 1/ 10

s = -1 gives D = 1/6

y = L^-1 { (-1/2) 1/(s+2) + (1/30) 1/(s-2) + (1/10) 1/(s+3) + (1/6) 1/(s+1)}

= -1/2 e^-2t + 1/30 e^2t + 1/10 e^-3t + 1/6 e^-t