Second order linear DE

  • Thread starter s7b
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s7b

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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

Homework Helper
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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 [itex]r=M \pm Ni[/itex]

then y=eMx(Acos(Nx)+Bsin(Nx))
 

s7b

26
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To find R1 and R2 do you just use the quadratic formula?

So r1= 1+i
and r2=1-i
 

rock.freak667

Homework Helper
6,230
31
To find R1 and R2 do you just use the quadratic formula?

So r1= 1+i
and r2=1-i
yep..so [itex]r= 1 \pm i[/itex] i.e. M=1 and N=1
 

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