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

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UnkleRhaukusBest ResponseYou've already chosen the best response.1
its 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 \[\big(x\mu(t)\big)'=q(t)\mu(t)\]
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
Here 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.
 one year ago

mathlillyBest ResponseYou've already chosen the best response.0
Thank you guys so much.
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
Generally 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 \(\mu=e^{\int P\ dx}\)
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
Oops, I meant \(y'+Py=Q\)
 one year ago
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