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Je m'appelle

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Ok, so I've been studying the method of variation of parameters in order to solve 2nd order ODEs, and I have a question regarding a supposition that is made in the definition of the method.

Say,

[tex]y'' + p(t)y' + q(t)y = g(t)[/tex]

Then the general solution to the above equation is

[tex]c_1y_1(t) + c_2y_2(t) + y_p [/tex], and now replacing [tex]c_1\ and\ c_2\ by\ u_1(t)\ and\ u_2(t)[/tex] on the complementary solution [tex]c_1y_1(t) + c_2y_2(t)[/tex] in order to find the particular solution [tex]y_p[/tex]

[tex]y_p = u_1(t)y_1(t) + u_2(t)y_2(t) [/tex]

[tex]y'_p = u'_1(t)y_1(t) + u_1(t)y'_1(t) + u'_2(t)y_2(t) + u_2(t)y'_2(t)[/tex]

And then the methods states that

[tex]u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0[/tex]

And this is where I'm curious and confused at the same time, I would like to know why it can be considered true, is there some kind of a proof to back this up? I mean, the person who developed this method certainly had something in mind to state that, so what does it consist of?

Then we get to derive [tex]y'_p[/tex] once again and then we substitute the values for [tex]y''_p,\ y'_p\ y_p[/tex] on the original equation to get the following system

[tex]u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0\ OBS:\ I\ wanna\ know\ why\ this\ is\ equal\ to\ zero.[/tex]

[tex]u'_1(t)y'_2(t) + u'_2(t)_2y'_2(t) = g(t) [/tex]

So we can solve for [tex]u_1(t)\ and\ u_2(t) [/tex] and find the particular solution for the ODE.

Leaving us with the general solution for the ODE

[tex]y = c_1y_1(t) + c_2y_2(t)\ -\ y_1(t)\int \frac{y_2(t)g(t)}{W(y_1,y_2)(t)}\ dt\ +\ y_2(t)\int \frac{y_1(t)g(t)}{W(y_1,y_2)(t)}\ dt [/tex]

Thanks.

Say,

[tex]y'' + p(t)y' + q(t)y = g(t)[/tex]

Then the general solution to the above equation is

[tex]c_1y_1(t) + c_2y_2(t) + y_p [/tex], and now replacing [tex]c_1\ and\ c_2\ by\ u_1(t)\ and\ u_2(t)[/tex] on the complementary solution [tex]c_1y_1(t) + c_2y_2(t)[/tex] in order to find the particular solution [tex]y_p[/tex]

[tex]y_p = u_1(t)y_1(t) + u_2(t)y_2(t) [/tex]

[tex]y'_p = u'_1(t)y_1(t) + u_1(t)y'_1(t) + u'_2(t)y_2(t) + u_2(t)y'_2(t)[/tex]

And then the methods states that

[tex]u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0[/tex]

And this is where I'm curious and confused at the same time, I would like to know why it can be considered true, is there some kind of a proof to back this up? I mean, the person who developed this method certainly had something in mind to state that, so what does it consist of?

Then we get to derive [tex]y'_p[/tex] once again and then we substitute the values for [tex]y''_p,\ y'_p\ y_p[/tex] on the original equation to get the following system

[tex]u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0\ OBS:\ I\ wanna\ know\ why\ this\ is\ equal\ to\ zero.[/tex]

[tex]u'_1(t)y'_2(t) + u'_2(t)_2y'_2(t) = g(t) [/tex]

So we can solve for [tex]u_1(t)\ and\ u_2(t) [/tex] and find the particular solution for the ODE.

Leaving us with the general solution for the ODE

[tex]y = c_1y_1(t) + c_2y_2(t)\ -\ y_1(t)\int \frac{y_2(t)g(t)}{W(y_1,y_2)(t)}\ dt\ +\ y_2(t)\int \frac{y_1(t)g(t)}{W(y_1,y_2)(t)}\ dt [/tex]

Thanks.

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