Solving a Second Order Linear Differential Equation

[s7b]

Hi,

I'm having problems solving this equation;

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

I found the auxiliary equation r^2 -2r +3 = 0
and since b^2 -4ac is less than zero this the case where r1 and r2 are complex numbers.

This is as far as I get without getting stuck.

Please help me out if you know this.
Thanks :)

 

[rock.freak667]

For ay''+by'+cy=0

the auxiliary equation is ar²+br+c=0. If b²-4ac <0, such that the roots are in the form r=M \pm Ni

then y=e^Mx(Acos(Nx)+Bsin(Nx))

 

[s7b]

To find R1 and R2 do you just use the quadratic formula?

So r1= 1+i
and r2=1-i

 

[rock.freak667]

To find R1 and R2 do you just use the quadratic formula?

So r1= 1+i
and r2=1-i

yep..so r= 1 \pm i i.e. M=1 and N=1