Solving Leplace Transform of y"+4y=7t with y(0)=-1, y'(0)=3

[hils0005]

Homework Statement

y"+4y=7t , y(0)=-1, y'(0)=3

The Attempt at a Solution

L{y} + 4L{y} = 7L{t}

(s^2L{y} - sy(0) - y'(0)) + 4L{y} = 7L{t}

s^2L{y} - s(-1) - 3 + 4L{y} = 7(0!/s^(0+1))

L{y}(s^2 + 4) = 7(0!/s^(0+1)) - s + 3
= ((-s+3)s + 7)/s
=(-s^2 + 3s +7)/s

L{y}=(-s^2 + 3s + 7)/(s(s^2+4))

here is where I run into trouble, setting up the partial fractions

L{y}=-s^2 + 3s +7 = A/s + B/(s^2+4)

is that correct so far?

 

[cheeee]

You should check the laplace transform for 7t.

L{t^n} = (n!)/(s^(n+1))
therefore L{t} = L{t^1} = (1!/(s^(1+1))

L{7t} should look more like 7*(1/s^2) or (7/s^2)

 

[hils0005]

but its t^1, so n=0, t^0+1=t^1 right?

 

[cheeee]

Its t^1, so n = 1 when you plug into the transform for t^n.

 

[sutupidmath]

Or if you want the Laplace transform of t^n in terms of gamma function:

$$L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}$$

$$\Gamma(n+1)=n\Gamma(n)=...=n!,n \in Z^+$$
$$\Gamma(1)=1$$