Answer:
Step-by-step explanation:
Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)
Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get
L{y'' - 6y' + 9y} = L{0} = 0
(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2
Using the theorem of the Laplace transform for derivatives, we know that:
Replacing the initial values y(0)=4, y′(0)=2 we obtain
and our differential equation (*) gets transformed in the algebraic equation
Solving for Y(s) we get
Now, we brake down the rational expression of Y(s) into partial fractions
The numerator of the addition at the right must be equal to 4s-22, so
A(s - 3) + B = 4s - 22
As - 3A + B = 4s - 22
we deduct from here
A = 4 and -3A + B = -22, so
A = 4 and B = -22 + 12 = -10
It means that
and
By taking the inverse Laplace transform on both sides and using the linearity of the inverse:
we know that
and for the first translation property of the inverse Laplace transform
and the solution of our differential equation is
