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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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    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.2012-12-05

1 Answers 1

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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