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mel0 Group Title

solve xdy/dx-y=-x

  • one year ago
  • one year ago

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  1. abb0t Group Title
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    First order linear differential equation. Put it into standard form first by dividing everything by x.

    • one year ago
  2. abb0t Group Title
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    You're first step is to find the integrating factor u(x)

    • one year ago
  3. abb0t Group Title
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    sorry p(x)*

    • one year ago
  4. abb0t Group Title
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    idk if you're there still or not to proceed explaining this...

    • one year ago
  5. abb0t Group Title
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    but your integrating factor is: \[e^{-\ln|x|} = \frac{ 1 }{ x }\] I'm going to skip the work for the next few steps, you can use it as a reference to absolutely check your work, but you SHOULD see that you get product rule as a result. Therefore: \[[\frac{ 1 }{ x }y]'=-1\] Next, take the integral: \[\int\limits [\frac{ 1 }{ x }y]'= - \int\limits dx = \frac{ 1 }{ x }y= -x+C\] \[y = -x^2+Cx\]

    • one year ago
  6. sirm3d Group Title
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    the equation should be \[\Large \frac{1}{x}y=-\int\frac{1}{x} (1) dx\]

    • one year ago
  7. abb0t Group Title
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    @sirm3d could you show him how you got that?

    • one year ago
  8. sirm3d Group Title
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    the differential equation is \[y'-\frac{1}{x}y=-1\] with integrating factor \(1/x\) the resulting equation is \[(1/x)y=\int(1/x)(-1)dx\]

    • one year ago
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