Differential Equations [a mere curiosity]

[iCeCuBEz v2]

It is known that the general equation of a homogeneous equation with constant coefficients that has an unrepeated pair of complex conjugate roots a plus or minus bi (with b not equal to 0) is as follows

e^ax(c1cosbx+c2sinbx)=y(x)

an example being

y''-4y'+5y=0

which has the characteristic equation

r^2-4r+5

which equals (by completing the square)

(r-2)^2+1=0

then solve for r in that characteristic equation yielding the complex conjugate roots 2 plus or minus i

then the general solution of the equation is

y(x)=e^(2x)(c1cosx+c2sinx)

I get how to do the procedure I just want to know why the general solution ends up being of the form e^ax(c1cosbx+c2sinbx)

[reuben_tate]

I haven't taken ODE's yet, but from the exposure I've had of them, I'd say the best way to prove the solution of an ODE is whatever it is to simply plug in the equation (y(x) = whatever) for the ODE it solves and verify.

However, this one is a bit more tricky to verify since it's not trivial what ODE a general solution solves. Let's try to see if we can derive it however. So we know the solution of the characteristic equation is a+bi and a-bi, so:

(r-(a+bi))(r-(a-bi)) = 0
r^2 - r(a+bi) - r(a-bi) + (a+bi)(a-bi) = 0
r^2 - r(2a) + (a^2 + b^2) = 0

Thus, the equation should solve the ODE:
y''-2ay'+(a^2+b^2)y = 0

We know:
y(x) = e^(ax)*(c1sin(bx)+c2cos(bx)) = c1e^(a+bi)x + c2e^(a-bi)x (Let me know if you need help understanding how I put it in this form)
y'(x) = c1(a+bi)*e^(a+bi)x + c2(a-bi)*e^(a-bi)x
y''(x) = c1(a+bi)^2*e(a+bi)x+c2(a-bi)^2*e^(a-bi)x

Plug and chug now. Gonna factor things from the get go:
((a+bi)^2 - 2a(a+bi) + (a^2+b^2))*c1*e^(a+bi)x +
((a-bi)^2 - 2a(a-bi)+ (a^2+b^2))*c2*e^(a-bi)x

=

(a^2+2abi -b^2 -2a^2-2abi+a^2+b^2)*c1*e^(a+bi)x +
(a^2-2abi -b^2 -2a^2+2abi +a^2+b^2)*c2*e^(a-bi)x

=

0 + 0 = 0 and BAM, we've verified it was indeed equal to zero. We have shown why the general solution solves it's corresponding ODE.

(Someone can correct me if I'm wrong, this is all my own derivation and I literally have no experience with this stuff, just my own insight xP)

EDIT: I literally had no idea where I was going with this when I started and I literally shat bricks when it worked out in the end (I was thinking I would just get stuck with some super long polynomial with complex bs thrown in which I wouldn't know what to do with). If my solution/derivation has some flaw, then oopsies, I'm sorry ;~; Otherwise, I'm glad I could help, it gives me hope that I'll be able to survive ODE's when I take it next semester P.S. Also thanks, now I know how to solve ODE's of this type :3

[Silvuh]

Originally Posted by reuben_tate

We know:
y(x) = e^(ax)*(c1sin(bx)+c2cos(bx)) = c1e^(a+bi)x + c2e^(a-bi)x (Let me know if you need help understanding how I put it in this form)

I think the question was why do we know this, so I'll expand on it.

When you start with the solution for unrepeated real roots for a homogenous equation with constant coefficients, e^(rt), you put your complex roots into r as usual, e^((λ±iμ)t).

Edit, a note on e^(rt): it makes sense to try using it as a solution for a second-order differential equation based on results from simple first-order ones. (Take the separable equation dy/dt = ry, which becomes dy/y = r*dt. Integrate both sides to get ln(y) = kt+c, raise to e, and you end up with y = c*e^(rt), exponential growth.)

Because we want to find real solutions to a real equation, we don't stop there. To do this, we use the Euler Formula e^(ix) = cos(x) +isin(x), which you can prove by looking at the series expansions of e^(ix), sin(x), and cos(x). (You can also make sense of it by looking at e^(ix) as rotation in the imaginary plane, which is why e^(iπ)=-1; follow the link if you'd like more explanation.)

e^((λ±ib)t) = e^(λt) * e^(±iμt) ____________ Addition/Subtraction of Exponents to Multiplication
__________ = e^(λt) * (cos(μt) ± isin(μt)) ___ Euler's Formula

y_1 = e^(λt)(cos(μt) + isin(μt))
y_2 = e^(λt)(cos(μt) - isin(μt))

Now thanks to the properties of linear combinations, we can make the following substitutions:

u_1 = (y_1 + y_2)/2
u_2 = (y_1 - y_2)/(2i)

Substitute and cancel, and you will get

u_1 = e^(λt)cos(μt) and u_2 = e^(λt)sin(μt)

which lead to the general solution:

y(t) = c_1*e^(λt)cos(μt) + c_2*e^(λt)sin(μt)

[reuben_tate]

Originally Posted by Silvuh

I think the question was why do we know this, so I'll expand on it.

Oops, good point. I kinda took that step as something "trivial" when I guess it isn't really trivial if it's something someone isn't familiar with, my bad.

Originally Posted by Silvuh

you put your complex roots into r as usual, e^((λ±iμ)t).

However, I took that the OP was wondering how do we know that the solutions of the characteristic equation of the ODE play a role in the solution of the ODE (gosh, I can't word anything tonight >.<) This isn't so trivial either :P

Thanks for the elaboration though on the part I didn't bother to elaborate on though, I'm sure it'll help with icecubez understanding

P.S. I kinda feel outta place here, I literally know next to nothing about ODE's ;~;

[Arkuski]

Yes yes, the imaginary roots translate into the series expansion of e^x. Of course, the solution for this type of ODE will be e^ax, but I'm guessing you need a better explanation as to why an imaginary root turns into something that has to do with sin and cos. Let's compute e^(a+bi). We can easily see that this is equal to (e^a)*(e^bi). So now, let's focus on e^bi. We plug this into the series expansion:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...

e^bi = 1 + (bi) + (bi)^2/2! + (bi)^3/3! + (bi)^4/4! + ...
= 1 + bi - b^2/2! - i(b^3) + b^4/4! + ...

We can separate even and odd terms in the series and find the following:

e^bi = (1 - b^2/2! + b^4/4! + ... ) + i(b - b^3/3! + b^5/5! + ... )

By recognizing these Taylor series, we can deduce the following:

e^bi = cos(b) + i*sin(b)

And, of course, we include our original e^a:

e^(a+bi) = (e^a) * [cos(b) + i*sin(b)]

[iCeCuBEz v2]

thanks for the explanation guys euler's formula is pretty legit

Arkusi that's fuckin sick