IVP using Laplace transform ?

For laplace transforms remeber that:

y' -> sY(s) -y(0)

Here y(0) is the value of y(t) when t=0.

Which becomes:

sY(s) -y(0) + Y(s) = 1/(s-4)

Y(s)[s+1]= 1/(s+4) + 2

Y(s) = 1/[(s-4)*(s+1)] + 2/(s+1)

Now you have to find partial frations of:

1/[(s-4)*(s+1)]

becomes:

(-1/5)*(1/(s-4)) + (1/5)*(1/(s+1))

So:

Y(s) = (-1/5)*(1/(s-4)) + (1/5)*(1/(s+1)) + 2/(s+1)

With some cleaning up:

Y(s) = (-1/5)*(1/(s-4)) + (11/5)*(1/(s+1))

Now you can find the inverse:

y(t) = (11/5)*e^-t + (-1/5)*e^4t

Partial fractions.. learn them love them breathe them.. Especially for laplace transforms.

The catch is in L(y')

Using Integration by Parts, you can show that L(y^(n)) (where n is the differentiation level, not an exponent) is equal to the series:

sY(s) - y^(n-1)(0) - s*y^(n-2)(0) - s^2*y^(n-3)(0) - ... - s^(n-2)*y'() - s^(n-1)*y(0)

Basically, this means that L(y') = sY(s) - y(0). Using this rule, your equation becomes:

L(y' + y) = L(e^4t)

sY(s) - 2 + Y(s) = 1/(s-4)

And no - you can't just throw y(0) onto one side of the equation to get it in there. This completely destroys the equation, and as my Professor would say, "All hope is lost."