Using Laplace Transforms to solve IVP's

dm59
Messages
5
Reaction score
0

Homework Statement


solve the ivp using laplace tranforms

y''+2y'+2y=0 y(0)=1 y'(0)=-3


Homework Equations





The Attempt at a Solution



get to Y(s)[s^2+2s+2]=s-1

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

^^^
don't know how to simplify the denominator to solve using Laplace transforms. If I had to guess I would say maybe partial fractions but keep getting the wrong answer when I try to use them.
 
Physics news on Phys.org
You've pretty much finished it, all you need to recognize that

Y(s)=\frac{s-1}{(s-1)^2+1}

through completing the square in the denominator. Now can you get that to work with

f(t)=L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2} \right\} = e^{at}\cos{(bt)}
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top