- #1

- 140

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Terilien
- Start date

- #1

- 140

- 0

- #2

- 66

- 0

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?

Which equation are you concerned with?

Last edited:

- #3

- 140

- 0

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.

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

Last edited:

- #4

- 66

- 0

Let's say your equation is

[tex]y^{\prime \prime} + \alpha(t) y^{\prime} + \beta(t) y = 0[/tex]

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

[tex]y(t) = e^{r(t)} [/tex]

bung this into the first equation, and you get

[tex]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[/tex]

Now, divide out the expoenents, and your equation becomes

[tex]r^{\prime \prime} + r^{\prime 2}+\alpha(t) r^{\prime} +\beta(t) =0[/tex]

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 [tex]q(t) =r^{\prime}(t)[/tex].

[tex]y^{\prime \prime} + \alpha(t) y^{\prime} + \beta(t) y = 0[/tex]

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

[tex]y(t) = e^{r(t)} [/tex]

bung this into the first equation, and you get

[tex]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[/tex]

Now, divide out the expoenents, and your equation becomes

[tex]r^{\prime \prime} + r^{\prime 2}+\alpha(t) r^{\prime} +\beta(t) =0[/tex]

So, you now have a

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

Last edited:

- #5

- 66

- 0

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

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

- #7

- 367

- 0

Is general recipe ,but sometimes such eqs can be solved explicitely and in finite ,closed form.

It depends on [itex]\alpha(t),\beta(t)[/itex] functions coefficients involved.

- #8

- 11

- 0

- #9

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

Yes, particularly "Euler type" or "equi-potential" equations.Is general recipe ,but sometimes such eqs can be solved explicitely and in finite ,closed form.

It depends on [itex]\alpha(t),\beta(t)[/itex] functions coefficients involved.

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

- #10

- 236

- 0

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...

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:

Share:

- Replies
- 8

- Views
- 7K

- Replies
- 6

- Views
- 10K

- Replies
- 1

- Views
- 1K