## differential equation problem

i'm trying to find what the intermediate steps were used in solving this differential equation:

$$y'' + y' -3y =0$$

$$m=\frac{-1\pm\sqrt{13}}{2}$$ now we have

$$y=C_1e^{\frac{-.5+\sqrt{13}}{2}}t + C_2e^{\frac{-.5 -\sqrt{13}}{2}}t$$

note that the .5 are 1\2. i couldn't get the latex for this to work
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 Recognitions: Gold Member Homework Help Science Advisor Do you know where "m" came from?

 Quote by arildno Do you know where "m" came from?

that m came from the auxillary equation of the differential
$$r^2 +r -3=0$$

Recognitions:
Gold Member
Homework Help

## differential equation problem

So what's bugging you, then?

Recognitions:
Gold Member
Homework Help
 Quote by RadiationX i'm trying to find what the intermediate steps were used in solving this differential equation: $$y'' + y' -3y =0$$ $$m=\frac{-1\pm\sqrt{13}}{2}$$ now we have $$y=C_1e^{\frac{-.5+\sqrt{13}}{2}}t + C_2e^{\frac{-.5 -\sqrt{13}}{2}}t$$ note that the .5 are 1\2. i couldn't get the latex for this to work
$$y=C_1e^{(-.5 + \frac{\sqrt{13}}{2})t} + C_2e^{(-.5-\frac{\sqrt{13}}{2})t}$$