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gfd43tg
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Homework Statement
Solve the following ordinary differential equations, where ##\alpha##, ##\beta##, and ##\lambda## are constants.
[tex]\frac {dy}{dx} + \beta y = 0 [/tex]
[tex]\frac {dy}{dx} + \beta y + \alpha = 0[/tex]
[tex] \frac {d^2y}{dx^2} - \lambda^2 y = 0[/tex]
[tex] \frac {d^2y}{dx^2} + \lambda^2 y = 0[/tex]
[tex] \frac {d^2y}{dx^2} - \lambda^2 y + \alpha = 0[/tex]
Homework Equations
The Attempt at a Solution
It has been about two years since I've had to solve anything other than non-homogeneous, separable differential equations, so I am pretty rusty on this. I know I put 5 questions in one thread, but since this is an exercise just to make me remember some stuff I learned a couple years ago and long forgot, I am consolidating them here as to not have to post a few threads.
1.
[tex]\frac {dy}{dx} + \beta y = 0 [/tex]
[tex] \int \frac {dy}{y} = \int -\beta dx [/tex]
[tex] ln |y| = -\beta x + C [/tex]
Assume y > 0
##y = e^{-\beta x + c}##
## y = Ce^{-\beta x}##
2.
[tex] \frac {dy}{dx} + \beta y + \alpha = 0 [/tex]
for this one, I'm not sure since it is non-homogeneous.
3.
[tex] \frac {d^2y}{dx^2} - \lambda^2 y = 0[/tex]
For this one,
[tex] \frac {d^2y}{dx^2} = \lambda^2y[/tex]
[tex]\frac {d}{dx} \frac {dy}{dx} = \lambda^2y[/tex]
From here I have an idea but I'm unsure this is legitimate math operations. How would I go about this one? I can see some similarity with problem 1
4.
[tex] \frac {d^2y}{dx^2} + \lambda^2 y = 0[/tex]
I believe this one has some sort of known solution, although I wouldn't know how to get it
##y = Acos(\lambda x) + Bsin(\lambda x)##
5.
[tex] \frac {d^2y}{dx^2} - \lambda^2 y + \alpha = 0[/tex]
once again non-homogeneous, and 2nd order nonetheless, so not sure what to do
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