What is the Solution to the Differential Equation y'' + y' -3y =0?

[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

 

[arildno]

Do you know where "m" came from?

 

[RadiationX]

Do you know where "m" came from?

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

 

[arildno]

So what's bugging you, then?

 

[BobG]

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

As you noted, you created the auxiliary (or characteristic) equation. The auxiliary equation was solved via the quadratic formula.

Note: You should have:
y=C_1e^{(-.5 + \frac{\sqrt{13}}{2})t} + C_2e^{(-.5-\frac{\sqrt{13}}{2})t}

 

[HallsofIvy]

If you were wondering where the characteristic equation came from, assume a solution ofthe form y= e^rt, plug it into the differential equation and see what r must be in order to make the equation true.