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Check DE general solution with complex roots

  • Thread starter boneill3
  • Start date
  • #1
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Homework Statement



Solve [itex]y''+4y'+5y=0[/itex]

find solutions for y(0)=1 and y'(0)=0

Homework Equations



Quadratic equation

The Attempt at a Solution



Hows this look ?

assume solution is in the form of [itex]y=ce^{rx}[/itex]

substitute [itex]y=ce^{rx}[/itex] into the equation.

[itex]cr^2e^{rx}+4cre^{rx}+5ce^{rx}=0[/itex]

we then divide by [itex]ce^{rx}[/itex] to give

[itex]r^2+4r+5=0[/itex]

So we than find the roots which are [itex]r_1=-2+i,r_2=-2-i[/itex]

which gives the solution of

[itex]y=c_1e^{-2+ix}+c_2e^{-2-ix}[/itex].

but as they are complex roots we use the equation

[itex]y=c_1e^{\lambda x}cos \mu x +c_2e^{\lambda x}sin \mu x }[/itex].


where [itex]\mu = 1 , \lambda = -2[/itex]

so

[itex]y(x)=c_1e^{-2x}cos x +c_2e^{-2x}sin x [/itex].

and

[itex]y'(x)=e^{-2x}(sin (x)(-c_1-sc_2)+cos(x)(c_2-2c_1)) [/itex].


to find [itex]c_1[/itex] and [itex]c_2[/itex] apply the initial conditions.

[itex]y(0) = c_1 = 1[/itex]
and
[itex]y'(0) = c_2-2= 0 [/itex] so [itex]c_2 = 2[/itex]

so that makes the general solution

[itex]y(x)=e^{-2x}cos x +2e^{-2x}sin x [/itex].
 
Last edited:

Answers and Replies

  • #2
33,636
5,296
It's easy to check.
1) Verify that y(0) = 1 and y'(0) = 2
2) Verify that your solution satisfies y'' + 4y' + 5y = 0.
 
  • #3
111
0
It's easy to check.
2) Verify that your solution satisfies y'' + 4y' + 5y = 0.

I got
y'' = -5
y' = 0
y =1

so its the solution thanks alot
 

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