Are My Solutions for 2nd Order ODE Real-Valued?

[amcavoy]

2nd Order ODE

y''+y=0

I come up with the solutions of y₁=c₁e^ix, y₂=c₂e^-ix. Now, using these I try to find the cooresponding real-valued solutions:

e^ix=cos(x)+isin(x)

e^-ix=cos(x)-isin(x)

Both of which have real-valued solutions cos(x). However, when I looked this up online, Wikipedia stated that the answer was y=c₁sin(x)+c₂cos(x). Am I doing something wrong here? Thanks.

 

[stunner5000pt]

both y_{1} = c_{1} e^{ix} and y_{2} = c_{2} e^{-ix} are solutions
Also when you add them up that too is a solution.
So when you add both the Y1 and Y2 solutions you should get the solution on wikipedia

 

[Integral]

y''+y=0

I come up with the solutions of y₁=c₁e^ix, y₂=c₂e^-ix. Now, using these I try to find the corresponding real-valued solutions:

e^ix=cos(x)+isin(x)

e^-ix=cos(x)-Iain(x)

What you are stating here is Euler's Identity, it is not derived from the solution of the differential equation but is the starting point for going from the imaginary exponential solution to the real valued trig solution.

Both of which have real-valued solutions cos(x). However, when I looked this up online, Wikipedia stated that the answer was y=c₁sin(x)+c₂cos(x). Am I doing something wrong here? Thanks.

All that separates you from the trig solution is Euler's and some algebra. Realize that the constants in the final solution will not be the same as the constants in your original solution.

 

[amcavoy]

both y_{1} = c_{1} e^{ix} and y_{2} = c_{2} e^{-ix} are solutions
Also when you add them up that too is a solution.
So when you add both the Y1 and Y2 solutions you should get the solution on wikipedia

Unfortunately I don't. Adding those together gives the following:

c_1e^{ix}+c_2e^{-ix}=\left(c_1+c_2\right)\cos{x}+i\left(c_1-c_2\right)\sin{x}=A\cos{x}+Bi\sin{x}

...which is not a real solution. Any ideas?

Thanks again.

 

[lurflurf]

if
y=a*exp(i*x)+b*exp(-i*x)
let
c=(a+b)/2
d=(a-b)/(2*i)
then
y=c*cos(x)+d*sin(x)
if you presume x real
y=Re[y]+i*Im[y]
={(Re[a]+Re)*cos(x)-(Im[a]sin(x)-Re)*sin(x)}+i*{(Im[a]+Im)*cos(x)+(Re[a]-Re)*sin(x)}
so for Im[y]=0 (real solution) it is required that
Im[a]+Im=0
Re[a]-Re=0
which is clearly equivalent to
Im[a+b]=0
Re[a-b]=0

 

[Hurkyl]

c_1e^{ix}+c_2e^{-ix}=\left(c_1+c_2\right)\cos{x}+i\left(c_1-c_2\right)\sin{x}=A\cos{x}+Bi\sin{x}

It looks like a real solution to me... (if you choose A and B properly...)

 

[amcavoy]

It looks like a real solution to me... (if you choose A and B properly...)

This seems strange to me. Let's say that B=k/i, which would cancel out the i in front of the sine. But now the coefficient isn't a real number. How can a solution be real if it contains complex numbers?

 

[lurflurf]

This seems strange to me. Let's say that B=k/i, which would cancel out the i in front of the sine. But now the coefficient isn't a real number. How can a solution be real if it contains complex numbers?

If the imaginary part is 0
ie the imaginary numbers cancel
is
1+i-i
real?
how about
i/i
e^(pi*i)
i^2
i^i
0*i
cos(i)
real?
2cos(pi/9) is a root of x^3-3x-1
the other roots are 2cos(7pi/9) and 2cos(13pi/9)
thus all are real.
The roots can be expressed in radicals, but only if complex numbers are used.

 

[Hurkyl]

To be a real function, all that is required is that when you plug in a real number, you get a real number out of it.