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anonymous
 3 years ago
How do you solve x'+3x = t + e^2t? Tried separation of variables, but didn't work
anonymous
 3 years ago
How do you solve x'+3x = t + e^2t? Tried separation of variables, but didn't work

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1its a linear equation ie, of the form \[x'+p(t)x=q(t)\] so find an integrating factor \[\mu(t)=e^{∫p(t)dt}\] then \[\big(x\mu(t)\big)'=q(t)\mu(t)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here we have a typical linear firstorder ordinary differential equation. The idea is that we observe \((\mu x)'=\mu x'+\mu'x\), so if we can make our lefthand side (\(x'+3x\)) look like \(\mu x'+\mu'x\), then we can integrate both sides to yield \(\mu x\) and therefore \(x\). We can do this using multiplication, so essentially we need to find some \(\mu\) such that \(\mu x'+3\mu x=\mu x'+\mu' x\), i.e. \(3\mu=\mu'\); it should be clear then that \(\mu=e^{3t}\). Multiply throughout to yield \(\mu x'+3\mu x=\mu t+\mu e^{2t}\), which is equivalent to \((\mu x)'=\mu t+\mu e^{2t}\). Integrate both sides and solve for \(x\) to yield \(x=\frac1\mu\int(\mu t+\mu e^{2t})dt\). Substitute in \(\mu=e^{3t}\) and simplify.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you guys so much.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Generally you can solve a linear firstorder ODE of the form \(y'+Px=Q\) where \(P,Q\) are functions of \(x\) as follows: \(y=\frac1\mu\int\mu Q\ dx\) where \(\mu=e^{\int P\ dx}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oops, I meant \(y'+Py=Q\)
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