# Approach of Variation of parameters (Variation of Constants)

i have a question about the approach "Variation of parameters (also known as variation of constants).

Imagine we have non-homogene ODE of the form: $$y' = a(x) \cdot y + b(x)$$ The homogene solution is found by the Eigenvalue approach or others like seperation of variables.

This leads us to: $$y_H = C \cdot \underbrace{e^{\int a(x) \mathop{dx}}}_{y_h} \qquad C \in \mathbb{C}$$

The next step to get the non-homogene solution is to use variation of constants with the approach: $$y_P = C(x)\cdot y_h$$ Why do we expect the special solution $$y_P$$ to be of the same kind as $$y_H$$? How do I know that the soluation must have the structure $$y(x) = y_H(x) + y_P(x)$$

I would greatly appreciate answers to the question.

For the first one, it's simply because someone figured out that that substitution leads to an equation of the form $$C'(x) = \cdots$$, which can be solved just by integrating it. I don't know who exactly was first for ODEs of order one, but the general method of variation of constants is credited to Euler and Lagrange.
(By the way, $$y_P = C \cdot y_H$$ doesn't need to look anything like $$y_H$$, since $$C$$ can be arbitrary complicated, so it's a bit of a stretch in general to say that they are “of the same kind”. But in simple cases, $$C$$ will be a simple function. It's not very far-fetched that if you have, for example, $$b(x)=\sin x$$ on the right-hand side, and $$a(x)$$ doesn't contain any trig functions, then you expect $$y_P$$ to be “some expression containing $$\sin x$$ and/or $$\cos x$$”, since how would otherwise that $$\sin x$$ appear when combining $$y_P$$ and $$y_P'$$?)
The second one (regarding $$y_H+y_P$$) is as very standard fact about linear equations (not only linear differential equations) which is explained in every textbook and surely many times on this site already; see here for one question about the case of second-order linear ODEs.
• Does applying to linear equations mean that this equation ($y = y_H + y_P$) is only true for linear differential equations? – Sebi2020 Dec 30 '18 at 20:09
• Did you mean $y_P = C \cdot y_H$ or did you mean $y_P = C \cdot y_h$? Many textbooks write $y_P = C \cdot y_H$ but i think that's not correct because of the earlier definition of $y_H$. This would evaluate to $y_P = C(x) \cdot C \cdot e^{\int a(x) dx}$. Or is it correct because you could substitute $C \cdot C(x)$ with $C^*(x) = C \cdot C(x)$ – Sebi2020 Dec 30 '18 at 20:12
• Yes, it's definitely only true for linear equations! And I should have written $y_h$, I didn't notice that notational distinction until now. But it really doesn't matter, because $y_H$ and $y_h$ only differ by a constant, and that constant can be absorbed into the function $C(x)$. – Hans Lundmark Dec 30 '18 at 20:17