# Second order homogenous with variable coeffecients

1. Mar 15, 2007

### Terilien

In general how do we deal with linear second order differential equations with variable coeffecients?

2. Mar 16, 2007

### Matthew Rodman

In general, you can't. You have to specify the form of those coefficients and hope to match it to an equation for which there is a known solution, probably involving a substitution or two along the way. (Unless it lends itself to direct integration, or something.)

Which equation are you concerned with?

Last edited: Mar 16, 2007
3. Mar 16, 2007

### Terilien

why can't you just plug in in e^rt like you do with the constant coefficient ones and just let r be variable(guessing)? For example when you plug e^rt into the homogenous equation you can find r by solving a quadratic equation involving the coffecient. You need to solve for r. however since in this case the coeeficients depend on time, can't we just have variable solutions (r). note that I haven't thought about this much.

I was just curious as the notes I found didn't discuss them much.

Last edited: Mar 16, 2007
4. Mar 16, 2007

### Matthew Rodman

Let's say your equation is

$$y^{\prime \prime} + \alpha(t) y^{\prime} + \beta(t) y = 0$$

where alpha and beta are arbitrary functions. Now, try your method of subbing

$$y(t) = e^{r(t)}$$

bung this into the first equation, and you get

$$r^{\prime \prime} e^{r(t)} + r^{\prime 2} e^{r(t)} + \alpha(t) r^{\prime} e^{r(t)} + \beta(t) e^{r(t)} = 0$$

Now, divide out the expoenents, and your equation becomes

$$r^{\prime \prime} + r^{\prime 2}+\alpha(t) r^{\prime} +\beta(t) =0$$

So, you now have a different equation, but, more than likely it's not one which is any easier to solve-- unless you get lucky.

Note however, this new one is effectively a first order equation -- i.e. by letting $$q(t) =r^{\prime}(t)$$.

Last edited: Mar 16, 2007
5. Mar 16, 2007

### Matthew Rodman

Err... if you want ot find out what kind of second-order equations are soluble, you can look here. They also have some solutions of PDEs on other pages.

6. Mar 17, 2007

### HallsofIvy

Staff Emeritus
Typically, series solutions are used for linear differential equations with variable coefficients.

7. Mar 17, 2007

### tehno

decomposition in series...

Is general recipe ,but sometimes such eqs can be solved explicitely and in finite ,closed form.
It depends on $\alpha(t),\beta(t)$ functions coefficients involved.

8. Mar 17, 2007

### Elvex

The brute force method, usually a method of last resort is the method of frobenius. The problem is you'll generate infinite series solutions which rarely have a closed form. The method is necessary for laplaces equation in cylindrical and spherical coordinates.

9. Mar 18, 2007

### HallsofIvy

Staff Emeritus
Yes, particularly "Euler type" or "equi-potential" equations.

Strictly speaking, "Frobenius" method only applies to series expansion about regular singular points, not general series expansions.

10. Mar 18, 2007

### ziad1985

There is the y(x)=u(x)*v(x)...
solve for v(x), and end up with something like this u'' +(something)u=0
if your lucky enough it's an easy equation...
Edit:If the Diff Eq was this: y'' +a(x)y'+b(x)y=0
then replacing y=v(x)*u(x)
we have this :u''*v+u'(2v'+a(x)v)+u(v"+a(x)v'+b(x)v)=0
Take (2v'+a(x)v)=0 solve for v , then replace in the Eq above.
Then this is what I meant if you were lucky (v"+a(x)v'+b(x)v) should simplify into something related to v in order to solve...

Last edited: Mar 18, 2007
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