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.


Those are two rather unrelated questions.

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.

  • $\begingroup$ Does applying to linear equations mean that this equation ($y = y_H + y_P$) is only true for linear differential equations? $\endgroup$ – Sebi2020 Dec 30 '18 at 20:09
  • $\begingroup$ 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 $C^*(x) = C \cdot C(x)$ $\endgroup$ – Sebi2020 Dec 30 '18 at 20:12
  • $\begingroup$ 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 $C(x)$. $\endgroup$ – Hans Lundmark Dec 30 '18 at 20:17

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