Solving Linear Homogeneous ODEs with Variable Coefficients

[foxjwill]

Homework Statement

Is there a standard method for finding the general solution to any given linear homogeneous ODE?

I tried working it out myself. Here's what I tried. I'm almost positive it's wrong because it doesn't take into account most of the coefficients. But maybe I'm on the right track?

The Attempt at a Solution

$$a_0(x)y + a_1(x)y' + \cdots + a_{n-1}(x)y^{(n-1)} + y^{(n)} = b(x)\qquad\qquad\qquad(1)$$​

Now, recall that the product rule for derivatives of order $$n$$ is

$$\frac{d^n}{dx^n}[f(x)g(x)] = {\sum^n_{k=0} \binom{n}{k} f^{(k)}(x)g^{(n-k)}}(x).$$​

We want to find some function $$u(x)$$ that, when multiplied to both sides of $$(1)$$, will put the left side in the form of the $$n$$th degree product rule. In other words,

$$u(x)a_0(x)y + u(x)a_1(x)y' + \cdots + u(x)y^{(n)} = \frac{d^n}{dx^n}[u(x)y]\qquad\qquad\qquad(2)$$​

After expanding $$(2)$$, we can set each term of the left side equal to the corresponding term on the right (e.g. $$u(x)a_2(x) y' = u^{(n-1)}(x)y'$$ ) and, after canceling out the all the $$y$$'s, form a system. Also note that since $$u(x)y^{(n)}$$ appears on both sides of $$(2)$$ it cancels out. From here until I say otherwise, all functions will be referenced, for convenience, without the "$$(x)$$" appended to it.

$$\left \{<br /> \begin{array}{lcl}<br /> u^{(n)} &=& u a_0\\<br /> u^{(n-1)} &=& \dbinom{n}{1} u a_1\\<br /> & \vdots\\<br /> u'' &=& \dbinom{n}{n-2} u a_{n-2}\\<br /> u' &=& \dbinom{n}{n-1} u a_{n-1}\\<br /> \end{array}<br /> \right .<br /> \qquad\qquad\qquad(3)$$​

Rearranging $$(3)$$, we have

$$\left \{<br /> \begin{array}{lcl}<br /> \dfrac{u^{n}}{u^{(n-1)}} &=& \dfrac{a_1}{a_0} \\<br /> \\<br /> \dfrac{u^{(n-1)}}{u^{(n-2)}} &=& \dbinom{n}{1} \dfrac{a_2}{a_1}\\<br /> & \vdots\\<br /> \dfrac{u'''}{u''} &=& \dbinom{n}{n-2} \dfrac{a_{n-2}}{a_{n-3}}\\<br /> \\<br /> \dfrac{u''}{u'} &=& \dbinom{n}{n-1} \dfrac{a_{n-1}}{a_{n-2}}\\<br /> \end{array}<br /> \right .<br /> \qquad\qquad\qquad(4)$$​

Solving each equation in $$(4)$$, we get

$$u^{(k)} = e^{\binom{n}{n-k} {\int \frac{a_{n-k}}{a_{n-(k+1)}}dx }}$$.​

Setting $$k=0$$,

$$u(x) = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}\qquad\qquad\qquad(5)$$​

Now, going back to $$(1)$$ and $$(2)$$, since, by definition,

$$\frac{d^n}{dx^n}[u(x)y] = u(x)b(x),$$​

we can plug in $$(5)$$,

$$\frac{d^n}{dx^n}\left[ e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx} y \right] = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}b(x)\qquad\qquad\qquad(6)$$​

and then antidifferentiate, which, at the moment, I am feeling too lazy to actually do.

 

[rock.freak667]

Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?

 

[foxjwill]

Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?

Yes, but I was wondering about methods to solve general degree ODE's with variable coefficients.

 

[Rainbow Child]

You have introduced one function $$u(x)$$ and you also have n equations in equation (3).

In general your solution for $$k=0$$, i.e. equation (5), does not satisfy all members of equation (3).
Thus there is no $$u(x)$$ in general.

 

[HallsofIvy]

Series solutions (including Frobenius' method) are the most general methods of solving linear equations.

 

[foxjwill]

You have introduced one function $$u(x)$$ and you also have n equations in equation (3).

In general your solution for $$k=0$$, i.e. equation (5), does not satisfy all members of equation (3).
Thus there is no $$u(x)$$ in general.

Yeah. I had a feeling it was something like that.

 

[foxjwill]

Series solutions (including Frobenius' method) are the most general methods of solving linear equations.

Hm. Darn. I don't like using series solutions. They're so cumbersome. Ah, well.

Also, I'm assuming the laplace and Fourier transforms, in general, can't be used to solve linear ODE's with variable coefficients? I know that there are a few specific cases where they work, but, yeah.