# Thread: "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

1. ## "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

I'm trying to "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms) and I realize my first mistake was where I put the red in the first attached document.
(The second attached document is the continuation of my incorrect work just for the sake of completeness).

How do I get to the final answer
y(x) = e^[(3/2) x] cos(sqrt(7)/2 x) + sqrt(7) e^[(3/2) x] sin(sqrt(7)/2 x)
from my last correct step?

(My book doesn't show the steps to get from where I'm stuck to the final answer).

Any help would be greatly appreciated!

2. ## Re: "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

I don't think you took the LT correctly. I get the following:

$\displaystyle s^{2}Y(s)-s y(0)-y'(0)-3(s Y(s)-y(0))+4Y(s)=0,$ or

$\displaystyle s^{2}Y(s)-s-5-3s Y(s)+3+4Y(s)=0,$ or

$\displaystyle s^{2}Y(s)-s-2-3s Y(s)+4Y(s)=0,$ or

$\displaystyle (s^{2}-3s+4)Y(s)=s+2.$

Can you continue from there?

3. ## Re: "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

Thank you for spotting my mistake with the Laplace Transform. I would have been able to continue if the two solutions for s^2 - 3s + 4 were real numbers but they are complex/imaginary numbers so I don't know how to continue on with the partial fractions step to break down Y(s) into fractions which are easy to inverse using a table of Laplace Transforms.

Did I make another mistake? If so, tell me please but if not, could you help me see what changes now that the solutions for s from the quadratic formula are s = [3 +/- sqrt(7) i]/2 ?

Edit: I'm attaching my handwritten work so far.

4. ## Re: "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

Express

$\displaystyle \frac{s+2}{s^2-3s+4}=\frac{s+2}{(s-3/2)^2+7/4}$

and use

$\displaystyle \mathcal{L}\{e^{-at}\cos bt\}=\frac{s+a}{(s+a)^2+b^2}\;,\quad \mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a)^2+b^2}$ .

5. ## Re: "Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

Thank you both. I was trying so hard to make it a product on the denominator. I didn't know I could make it a sum and use the table of Laplace Transforms successfully like that.

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# y''' 3y''-4y:0

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