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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.

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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.

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  • $\begingroup$ 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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