- #1
twoflower
- 368
- 0
Hi,
I just started playing with higher order ODEs and I'm stuck in one particular step. Here it is:
[tex]
y^{''} + y = \frac{1}{\cos x}
[/tex]
1. step: I find fundamental solution system, which in this case is
[tex]
[\cos x, \sin x]
[/tex]
So general solution looks like this:
[tex]
y(x) = \alpha\cos x + \beta \sin x
[/tex]
Using the method of variation of parameters, [itex]\alpha[/itex] and [itex]\beta[/itex] become functions of x:
[tex]
y(x) = \alpha(x)\cos x + \beta(x) \sin x
[/tex]
[tex]
y'(x) = \alpha^{'}(x)\cos x - \alpha(x)\sin x + \beta^{'}(x) \sin x + \beta(x) \cos x
[/tex]
Now I don't understand the condition
[tex]
\alpha^{'}(x)\cos x + \beta^{'}(x) \sin x = 0
[/tex]
Why does it have to be so?
Thanks for explanation!
I just started playing with higher order ODEs and I'm stuck in one particular step. Here it is:
[tex]
y^{''} + y = \frac{1}{\cos x}
[/tex]
1. step: I find fundamental solution system, which in this case is
[tex]
[\cos x, \sin x]
[/tex]
So general solution looks like this:
[tex]
y(x) = \alpha\cos x + \beta \sin x
[/tex]
Using the method of variation of parameters, [itex]\alpha[/itex] and [itex]\beta[/itex] become functions of x:
[tex]
y(x) = \alpha(x)\cos x + \beta(x) \sin x
[/tex]
[tex]
y'(x) = \alpha^{'}(x)\cos x - \alpha(x)\sin x + \beta^{'}(x) \sin x + \beta(x) \cos x
[/tex]
Now I don't understand the condition
[tex]
\alpha^{'}(x)\cos x + \beta^{'}(x) \sin x = 0
[/tex]
Why does it have to be so?
Thanks for explanation!