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- Thread starter Terilien
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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?

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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.

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- #4

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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].

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- #6

HallsofIvy

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

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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.

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- #9

HallsofIvy

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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.

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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...

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