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mathlilly

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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  1. UnkleRhaukus
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    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
  2. oldrin.bataku
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    Here we have a typical linear first-order ordinary differential equation. The idea is that we observe \((\mu x)'=\mu x'+\mu'x\), so if we can make our left-hand 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
  3. mathlilly
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    Thank you guys so much.

    • one year ago
  4. oldrin.bataku
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    Generally you can solve a linear first-order 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
  5. oldrin.bataku
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    Oops, I meant \(y'+Py=Q\)

    • one year ago
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