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Say I have a general 1st-order homogeneous linear DE with constant coefficients.

$y'+ay=0$

The solution is $y=C_0e^{ax}$

Say I have the same thing but non-homogeneous.

$y'+ay=b$

The solution is $y=\frac{b}{a}+C_oe^{ax}$

The difference, obviously, is $\frac{b}{a}$. $\frac{b}{a}$ itself is a solution, and a constant solution at that. Bear with me, there's more.

So let's say that I don't have constant coefficients. Here's the homogeneous DE:

$y'+a(x)y=0$

The solution is $y=C_0e^{\int a(x)dx}$

And the non-homogeneous DE:

$y'+a(x)y=b(x)$

The solution being $y=e^{-\int a(x)dx} \int b(x)e^{\int a(x)dx}dx+C_0e^{\int a(x)dx}$

So the difference between the solutions is $e^{-\int a(x)dx} \int b(x)e^{\int a(x)dx}dx$.

So here's my question: Is there anything significant about each of the first terms in the solutions for both non-homogeneous solutions? They're both solutions themselves, but is there something particular about them, that separates them from all the other solutions?

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You might be asking something deeper -- but for linear ODE's (like the ones you mentioned), the general solution is always the sum of the general solution to the homogeneous problem plus one solution to the nonhomogeneous problem. –  StuartHa Dec 3 '12 at 6:05
    
I don't think these solutions are any more (or less) special than, say, $-\cos x$ is special as a value of $\int\sin x\,dx$. –  Gerry Myerson Dec 3 '12 at 6:27
    
@Stuart So the one solution to the non-homogeneous problem: can it be any solution? If I replace either of the first terms above with any other solution, would I still have the general solution to each problem? –  Korgan Rivera Dec 3 '12 at 14:07
    
Yes --- why not try it, and see? –  Gerry Myerson Dec 3 '12 at 23:10
    
Yeah it doesn't work. I thought there might be something particular about the solution but it doesn't seem like there is. It's just the solution when $C_0$ is zero, nothing more. –  Korgan Rivera Dec 5 '12 at 16:37

1 Answer 1

There is something particular about the solution. In both cases what you have is a equations of the form $$ \sum_{k=1}^n a_k(x) y^{k)}(x) = b(x) $$ this is a linear differential equation of first order. Let $$ L (y) = \sum_{k=1}^n a_k(x) y^{k)}$$ then $L$ is what is called a linear differential operator. What it does is it takes a n-times differentiable function and gives another function. You can check the property $$ L(u + v) = L(u) + L(v), \qquad L(\alpha u) = \alpha L(u), \ \alpha \in \mathbb R$$ (this is why is called linear). You you have in each problem is $Ly = b$ that is to find a function so that its image through $L$ is precisely $b$. If you take any functions $y_0, y_1$ such that $$ Ly_0 = 0, Ly_1 = b $$ then you have that $$ L(y_0 + y_1) = b $$ Since in both your problem one can readily solve the problem $$ Ly_0 = 0$$ Then one can consider what is called the variation of constant formulas. Which consists on assuming the existence of a solution for $Lu = b$ of the form $y_1(x) = A(x) y_0(x)$. Try to compute $L(A(x)y_0)$ in both of them and then see what you get.

If you are not able to check it you can always google variation of constant formulas.

Regards, D

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