Approach of Variation of parameters (Variation of Constants)
$begingroup$
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.
linear-algebra ordinary-differential-equations fundamental-solution
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra ordinary-differential-equations fundamental-solution
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra ordinary-differential-equations fundamental-solution
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
$endgroup$
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.
linear-algebra ordinary-differential-equations fundamental-solution
linear-algebra ordinary-differential-equations fundamental-solution
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
asked Dec 30 '18 at 17:34
Sebi2020Sebi2020
1225
1225
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
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.
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
answered Dec 30 '18 at 18:10
Hans LundmarkHans Lundmark
35.3k564114
35.3k564114
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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