Complex exponentials and differential equations

[Pixter]

guestion: Use complex exponentials to find the solution of the differential equation

(d^2y(t)/dt^2) + (3dy(t)/dt) + (25/4)y(t) = 0

such that y(0) = 0, dy/dt =1 for t=o

my taughts: I started by putting it in the form m^2 + 3m +25/4
m = (-3sqrt(9-25))/2 = (-3sqrt(-16))/2 = (-3+-4i)/2

then i thaught one can put it in the form e^pt(AcosQt+BsinQt) [p+-Qi]

so: y = (e^(-3/2)t)(Acos2t + Bsin2t)

y(0)=0 dy/dt=1 for t=0 y=Ae^(((-3+4i)/2)t) + Be^(((-3-4i)/2)t)

0 = (e^(-3/2)t)(Acos2t + Bsin2t)
0 = (Acos2t + Bsin2t)
0 = A + 0
A=0
dy/dt = (-3/2(e^(-3/2)t))(Acos2theta + Bsin2theta) + (e^(-3/2)t)(-2Asin2t + 2Bcos2t)
1=(-3/2)A +2B
2B = 1 (because A=0)
B=1/2
so y(t) = e^((-3/2)t) ((1/2)sin2t)



Don't know if I have done the question right or even got the question at all. just wanted to know if this is right, or if I'm on the right track but made a misstake on the way. Also if I'm completely wrong please point that out and give me a pointer where to start. Cheers

 

[HallsofIvy]

guestion: Use complex exponentials to find the solution of the differential equation
(d^2y(t)/dt^2) + (3dy(t)/dt) + (25/4)y(t) = 0
such that y(0) = 0, dy/dt =1 for t=o
my taughts: I started by putting it in the form m^2 + 3m +25/4
m = (-3sqrt(9-25))/2 = (-3sqrt(-16))/2 = (-3+-4i)/2
then i thaught one can put it in the form e^pt(AcosQt+BsinQt) [p+-Qi]
so: y = (e^(-3/2)t)(Acos2t + Bsin2t)
y(0)=0 dy/dt=1 for t=0 y=Ae^(((-3+4i)/2)t) + Be^(((-3-4i)/2)t)
0 = (e^(-3/2)t)(Acos2t + Bsin2t)
0 = (Acos2t + Bsin2t)
0 = A + 0
A=0
dy/dt = (-3/2(e^(-3/2)t))(Acos2theta + Bsin2theta) + (e^(-3/2)t)(-2Asin2t + 2Bcos2t)
1=(-3/2)A +2B
2B = 1 (because A=0)
B=1/2
so y(t) = e^((-3/2)t) ((1/2)sin2t)
Don't know if I have done the question right or even got the question at all. just wanted to know if this is right, or if I'm on the right track but made a misstake on the way. Also if I'm completely wrong please point that out and give me a pointer where to start. Cheers

It's not at all clear what you did! You apparently wrote the solution to the differential equation in terms of real functions, sine and cosine, converted those to complex exponential, found the constants A and B to give the initial conditions and then converted back to real functions. I don't believe that was what was intended by "use complex exponentials to find the solution".

First solve the characteristic equation m² + 3m +25/4= 0.
(I don't like your phrasing "putting it in the form". You didn't put the differential equation in this form- this is the characteristic equation for the differential equation. Also you didn't write it as an equation, you just wrote "m² + 3m +25/4" and then declared what m must be!)
You solved that correctly: m= (3/2)+ i and (3/2)- i.
Remember that one way of getting the characteristic equation is to look for a solution of the form e^mx. You don't need to go through the "sine", "cosine" form: the general solution to the differential equation, in terms of complex exponentials is:
$$y= C_1e^{(\frac{3}{2}+ i)t}+ C_2e^{(\frac{3}{2}-i)t}$$
Taking t= 0 in that gives $$y(0)= C_1+ C_2= 0$$
Differentiating the general solution gives
$$y'= (\frac{3}{2}+ i)C_1e^{(\frac{3}{2}+ i)t}+(\frac{3}{2}- i)C_2e^{(\frac{3}{2}+ i)t}$$
$$y'(0)= (\frac{3}{2}+ i)C_1+ (\frac{3}{2}+ i)C_2= 1$$.

Once you have found C₁ and C₂ I see no reason why you can't just leave the solution in "complex exponential" form!

 

[Architeuthis Dux]

Yes, you are on the right track! Your approach using complex exponentials is correct. However, there is a small mistake in your final equation for y(t). The correct solution is y(t) = e^(-3/2t)(1/2)sin(2t) + e^(-3/2t)(1/2)cos(2t). This can be verified by plugging it into the original differential equation and using the initial conditions. Keep up the good work!