DE and Complex numbers

1. Feb 17, 2005

sisigsarap

I believe that I have worked the problem correctly to the point where I am, however I am not sure how to incorporate complex numbers into my answer?

The problem is :Make the substitution v = ln x to solve 4x^2 * y" + 8xy' - 3y = 0. Where " represents double prime, and ' represents prime.

This is my work so far:
v = ln x dv/dx = 1/x

dy/dx = dy/dv * dv/dx
dy/dx = dy/dv * 1/x

d"y/dx" = d/dx [dy/dv * 1/x] - dy/dv * 1/x^2
= 1/x * d/dx[dy/dv] - dy/dv * 1/x^2
= 1/x^2 * d"y/dv" - dy/dv * 1/x^2

Then plugging into the original equation:

4x^2[t/x^2 * d"y/dv" - dy/dv * 1/x^2] + 8x[dy/dv * 1/x] - 3y = 0
which can be broken down to
4d"y/dv" + 4dy/dv - 3y = 0

Substituting in R, i get 4r^2 + 4r - 3 = 0
This does not readily look factorable to me so I use the quadratic equation.
[-4 +- Square Root of (16 - 48)]/8

Here is where I run into the problem, does that turn into [-2 +- Square root of (8i)] / 4

It has been a while since I have dealt with complex numbers and I do not recall how to manipulate them.

If that is the correct equation, can anyone tell me how to finish the problem?

Thanks,

Josh

2. Feb 17, 2005

dextercioby

Sure,use Euler's identity:
$$e^{i\theta}\equiv \cos\theta+i\sin\theta$$

for the cmplex exponential part.

Daniel.

3. Feb 17, 2005

sisigsarap

So would I be correct in saying that :

y = e^((-1/2)*x)*[(C1Cos((Square root of 2)/2)*X) + (C2Sin((Square root of 2)/2)*X)]

is correct?

Where C1 represents some constant and C2 represents some other constant.

4. Feb 17, 2005

dextercioby

Not "x",but "v"...So in terms of "x",u should have some sine & cosine of natural logarithms...

Daniel.

5. Feb 17, 2005

ehild

You've made a little mistake, it is 16+48 under the square root. You know, (-b+-sqrt(b^2-4ac))/2a. is the solution of the equation ax^2 + bx +c =0.

ehild

6. Feb 17, 2005

sisigsarap

Wow! Thanks for the help guys!

And thanks for catching my mistake ehild! I guess that is why I shouldnt work on these problems at the wee hours of the morning.

7. Feb 17, 2005

dextercioby

Daniel.

P.S.So you have a combination of sinh and cosh...

8. Feb 17, 2005

saltydog

This is a Euler equation in the standard form:

$$ax^my^{(m)}+bx^{(m-1)}y^{(m-1)}+...+ky=0$$

The general solution is of the form: $y(x)=\sum_i^m c_ix^{(m_i)}$
where $m_i$ are the roots of the polynomial when $x^m$ is substituted into the ODE.

The polynomial equation for the ODE of the original posts is:

$$4m^2+4m-3=0$$

Thus, the solution is seen to be:

$$y(x)=c_1x^{\frac{1}{2}}+c_2x^{\frac{-3}{2}}$$

Can you attempt to solve the following:

$$4x^2y''+8xy'+3y=0$$

This produces complex roots which necessarily involve terms of the form:
$\cos(\ln[u(x)])$ and $\sin(\ln[v(x)])[/tex] in which [itex]u(x)$ and $v(x)$ are powers of x. Just absorb the imaginary i into the arbritrary constants to get a real-valued function. Although I believe the function is valid only for x>0 because of the logarithm involved. This is what I got:
y[x]={c_1}\multsp \frac{\cos \big[\ln \big[{x^{\frac{1}{{\sqrt{2}}}}}\big]\big]}{{\sqrt{x}}}+\\ \noalign{\vspace{2.125ex}} \hspace{2.em} {c_2}\multsp \frac{\sin \big[\ln \big[{x^{\frac{1}{{\sqrt{2}}}}}\big]\big]}{{\sqrt{x}}}

Last edited: Feb 17, 2005