Mixing
why is it possible to make the extra assumption that $(x-a)h'(x)+(b-x)g'(x) = 0$
$begingroup$
In my book they study the differential equation :
$$y''+fy = 0$$
where $f$ is continuous on $[a,b]$, $y in C^2([a,b], mathbb{R})$ and such that the above equation has a solution $u$ with $u(a) = u(b) = 0$
Then they say the following :
$x-a, x-b$ is basis of the equation $y'' = 0$, hence we can write
$$u(x) = (x-a)h(x) + (b-x)g(x)$$
This makes sense to me, but then they say : with :
$$u(x) = (x-a)h'(x) + (b-x)g'(x) = 0$$
I don't understand why it's obvious that there exist two functions $h$ and $g$ that verify these two conditions. The first one is obvious to me, but I don't understand why it's possible to also have the second condition.
Thank you !
real-analysis calculus ordinary-differential-equations
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
In my book they study the differential equation :
$$y''+fy = 0$$
where $f$ is continuous on $[a,b]$, $y in C^2([a,b], mathbb{R})$ and such that the above equation has a solution $u$ with $u(a) = u(b) = 0$
Then they say the following :
$x-a, x-b$ is basis of the equation $y'' = 0$, hence we can write
$$u(x) = (x-a)h(x) + (b-x)g(x)$$
This makes sense to me, but then they say : with :
$$u(x) = (x-a)h'(x) + (b-x)g'(x) = 0$$
I don't understand why it's obvious that there exist two functions $h$ and $g$ that verify these two conditions. The first one is obvious to me, but I don't understand why it's possible to also have the second condition.
Thank you !
real-analysis calculus ordinary-differential-equations
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
1
$begingroup$
See "variation of constants" where the same reasoning is used. $u$ is one function that is parametrized by two functions, so you have one functional relation between $g$ and $h$ free. It can be used to reduce the order of derivatives in the computation of the second derivatives of $u$.
$endgroup$
– LutzL
Jan 26 at 11:36
add a comment |
$begingroup$
In my book they study the differential equation :
$$y''+fy = 0$$
where $f$ is continuous on $[a,b]$, $y in C^2([a,b], mathbb{R})$ and such that the above equation has a solution $u$ with $u(a) = u(b) = 0$
Then they say the following :
$x-a, x-b$ is basis of the equation $y'' = 0$, hence we can write
$$u(x) = (x-a)h(x) + (b-x)g(x)$$
This makes sense to me, but then they say : with :
$$u(x) = (x-a)h'(x) + (b-x)g'(x) = 0$$
I don't understand why it's obvious that there exist two functions $h$ and $g$ that verify these two conditions. The first one is obvious to me, but I don't understand why it's possible to also have the second condition.
Thank you !
real-analysis calculus ordinary-differential-equations
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
In my book they study the differential equation :
$$y''+fy = 0$$
where $f$ is continuous on $[a,b]$, $y in C^2([a,b], mathbb{R})$ and such that the above equation has a solution $u$ with $u(a) = u(b) = 0$
Then they say the following :
$x-a, x-b$ is basis of the equation $y'' = 0$, hence we can write
$$u(x) = (x-a)h(x) + (b-x)g(x)$$
This makes sense to me, but then they say : with :
$$u(x) = (x-a)h'(x) + (b-x)g'(x) = 0$$
I don't understand why it's obvious that there exist two functions $h$ and $g$ that verify these two conditions. The first one is obvious to me, but I don't understand why it's possible to also have the second condition.
Thank you !
real-analysis calculus ordinary-differential-equations
real-analysis calculus ordinary-differential-equations
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 10:49
dghkgfzyukzdghkgfzyukz
16612
16612
1
$begingroup$
See "variation of constants" where the same reasoning is used. $u$ is one function that is parametrized by two functions, so you have one functional relation between $g$ and $h$ free. It can be used to reduce the order of derivatives in the computation of the second derivatives of $u$.
$endgroup$
– LutzL
Jan 26 at 11:36
add a comment |
1
$begingroup$
See "variation of constants" where the same reasoning is used. $u$ is one function that is parametrized by two functions, so you have one functional relation between $g$ and $h$ free. It can be used to reduce the order of derivatives in the computation of the second derivatives of $u$.
$endgroup$
– LutzL
Jan 26 at 11:36
1
1
$begingroup$
See "variation of constants" where the same reasoning is used. $u$ is one function that is parametrized by two functions, so you have one functional relation between $g$ and $h$ free. It can be used to reduce the order of derivatives in the computation of the second derivatives of $u$.
$endgroup$
– LutzL
Jan 26 at 11:36
$begingroup$
See "variation of constants" where the same reasoning is used. $u$ is one function that is parametrized by two functions, so you have one functional relation between $g$ and $h$ free. It can be used to reduce the order of derivatives in the computation of the second derivatives of $u$.
$endgroup$
– LutzL
Jan 26 at 11:36
add a comment |
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$begingroup$
See "variation of constants" where the same reasoning is used. $u$ is one function that is parametrized by two functions, so you have one functional relation between $g$ and $h$ free. It can be used to reduce the order of derivatives in the computation of the second derivatives of $u$.
$endgroup$
– LutzL
Jan 26 at 11:36