Solve by the method of Laplace Transform y 2y 5y 3ex sinx

Solve by the method of Laplace Transform

y\'\' + 2y\' + 5y = 3e^-x sin(x) , y(0) = 0 , y0 (0) = 3

Please show all the steps when taking the laplace inverse.

Solution

Taking Laplace Transform of the given equation

L[ y\'\' ] + 2L[ y\' ] + 5L[ y ] = L[ 3e^-x sin(x) ]

s²Y - s.y(0) - y\'(0) + 2Y + 5Y = 3/( (s + 1)² + 1 )

Plugging the given values of y(0) = 0 and y\'(0) = 3

s²Y - 3 + 2sY + 5Y = 3/( (s + 1)² + 1 )

Y( s² + 2s + 5 ) = 3 + ( 3/( (s + 1)² + 1 ) )

Y = 1/( s² + 2s + 5 ) ( 3 + ( 3/( (s + 1)² + 1 ) ) )

Y = 3/( s² + 2s + 5 ) + 3/( ( s² + 2s + 5 )*( s² + 2s + 2 ) )

Y = 3/( s² + 2s + 5 ) + ( ( s² + 2s + 5 ) - ( s² + 2s + 2 ) )/( ( s² + 2s + 5 )*( s² + 2s + 2 ) )

Y = 3/( s² + 2s + 5 ) + 1/( s² + 2s + 2 ) - 1/( s² + 2s + 5 )

Y = 1/( s² + 2s + 2 ) + 2/( s² + 2s + 5 )

Y = 1/( (s+1)² + 1² ) + 2/( (s + 1)² + 2² )

Taking Inverse Laplace Transform

y(t) = e^-tsin(t) + e^-tsin(2t)

y(t) = e^-t( sin(t) + sin(2t) )