non-factorable 2nd order linear hom. ODE: y''+5y'+3y=0 y(0)=1 y'(0)=0 r2+5r+3=0

[aubs]

Hi, I am having trouble solving for the general solution of this DE:
y''+5y'+3y=0 y(0)=1 y'(0)=0
r²+5r+3=0
The part I'm struggling with is that I don't think this factorable and thus am not sure how to go about solving for the general solution. I do understand the proceeding steps.

 

[HallsofIvy]

Is "factoring" the only way you know to solve a quadratic equation? And you are taking differential equations?

You can also "complete the square":
\(\displaystyle x^2+ 5x= -3\)
\(\displaystyle x^2+ 5x+ 25/4= -3+ 25/4= 13/4\)
\(\displaystyle (x+ 5/2)^2= 13/4\)
\(\displaystyle x+ 5/2= \pm \sqrt{13}/2\).
\(\displaystyle x= -5/2\pm \sqrt{13}/2\).

You could also use the "quadratic formula": The solutions to \(\displaystyle ax^2+ bx+ c= 0\) are given by \(\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2}\). Here a= 1, b= 5, and c= 3 so the quadratic formula gives
\(\displaystyle x= \frac{-5\pm\sqrt{5^2- 4(1)(3)}}{2}= \frac{-5\pm\sqrt{13}}{2}\).