How Do You Integrate Complex Numbers into Differential Equations Solutions?

[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

 

[dextercioby]

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

for the cmplex exponential part.

Daniel.

 

[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.

 

[dextercioby]

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

Daniel.

 

[ehild]

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



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


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

Josh

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

 

[sisigsarap]

Wow! Thanks for the help guys!

And thanks for catching my mistake ehild! I guess that is why I shouldn't work on these problems at the wee hours of the morning.

 

[dextercioby]

Ups,i didn't follow your calculations...Just commented on the final result.

Daniel.

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

 

[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 [itex]\sin(\ln[v(x)])[/tex] in which $u(x)$ 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:
$$y[x]={c_1}\multsp \frac{\cos \big[\ln \big[{x^{\frac{1}{{\sqrt{2}}}}}\big]\big]}{{\sqrt{x}}}+\\ \noalign{\vspace{2.125ex}} \hspace{2.em} {c_2}\multsp \frac{\sin \big[\ln \big[{x^{\frac{1}{{\sqrt{2}}}}}\big]\big]}{{\sqrt{x}}}$$