2 Attachment(s)

"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!

Thanks in advance!

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?

1 Attachment(s)

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.

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} $ .

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.