Second order homogenous with variable coeffecients

[Terilien]

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

 

[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?

 

[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.

 

[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 $$q(t) =r^{\prime}(t)$$.

 

[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.

 

[HallsofIvy]

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

 

[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.

 

[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.

 

[HallsofIvy]

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.

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

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.

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

 

[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...