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Calculus and Beyond Homework Help
Solving First Order Linear DEs: Methods and General Solutions
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[QUOTE="BOAS, post: 5247845, member: 488860"] Hi, I have a first order linear DE that I need to find the general solution for. I thought that I had, but my solution does not make sense when plugged back into the equation. I think that my method of separation of variables might be inapplicable here, but don't know the reason for this. I know that for a DE of the form; [itex]y' + ay = b[/itex], the general solution is given by [itex]y = \frac{b}{a} + Ce^{-ax}[/itex] but I don't know where this result comes from, so I would greatly appreciate some help here. I will show what I have done, even though it is wrong. [B] 1. Homework Statement [/B] Find the general solution to; [itex]c(\phi) : c' + 2c = 1[/itex] [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] [/B] Rewrite as [itex]\frac{dc}{d \phi} = 1 - 2c[/itex] [itex]\frac{dc}{1 - 2c} = d\phi[/itex] Integrate both sides. [itex]\int \frac{dc}{1 - 2c} = \int d\phi[/itex] [itex]- \frac{1}{2} ln(|1 - 2c|) + a_{1} = \phi + a_{2}[/itex] Exponentiate, to get rid of ln, and let a_2 - a_1 = a [itex](1 - 2c)^{-\frac{1}{2}} = e^{\phi} + e^{a}[/itex] [itex]c = \frac{1}{2} - \frac{e^{-2\phi} + e^{-2a}}{2}[/itex] which is not a solution to my differential equation... [itex]c = \frac{1}{2} + Qe^{-2 \phi}[/itex] is. (Using Q as my constant to avoid confusion) So I have two questions; 1) Why can I not use separation of variables? (Or did I make a mistake?) 2) What is the argument that leads to the aforementioned 'formula' for solutions? Thanks. [/QUOTE]
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Solving First Order Linear DEs: Methods and General Solutions
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