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DanielM_113 Group Title

What is the general solution to the ODE /[dy/dx = 2y+1?/] Use separation of variables. Is y' + xy = x separable?

  • one year ago
  • one year ago

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  1. DanielM_113 Group Title
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    \[\int \frac{dy}{2y+1}=\int dx \\ \ 2 \ln y = x \]

    • one year ago
  2. DanielM_113 Group Title
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    What are the steps to the integration of the equation? \[\int \frac{dy}{2y+1} \\ \frac{dt}{dy} = 2y+1 \\ \int dt = \int\\ \int \frac{dt}{t} = \ln t \\ \]

    • one year ago
  3. DanielM_113 Group Title
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    Hmm, Im having trouble remembering this.

    • one year ago
  4. DanielM_113 Group Title
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    \[t=2y+1 \\ \frac{1}{t}=\frac{1}{2y+1} \\ \frac{dt}{dy}=2 \\\] Is this true? \[\int\frac{dt}{t} = \int \frac{dy}{2y+1}\quad\mbox{?} \]

    • one year ago
  5. DanielM_113 Group Title
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    I need to visualize integration by substitution better.

    • one year ago
  6. DanielM_113 Group Title
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    if I set \[u=2y+1\\ \frac{\mathrm du}{\mathrm dy}= 2\\ \frac{1}{u}\frac{\mathrm du}{\mathrm dy}=\frac{1}{2y+1}2 \\ \frac{1}{2}\int \frac{\mathrm du}{u}=\int\frac{1}{2y+1}\mathrm dy \\ \int \frac{\mathrm du}{u} = \ln u = \ln (2y +1)\\ \int\frac{1}{2y+1}\mathrm dy =\frac{1}{2}\int \frac{\mathrm du}{u}=\frac{1}{2}\ln (2y +1) \]

    • one year ago
  7. DanielM_113 Group Title
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    \[\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 -c_1\\ 2y+1=e^{2x+2c_2-c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}e^{c_3}e^{2x}\\ y=\frac{1}{2}Ce^{2x} \] This looks right for the first question.

    • one year ago
  8. DanielM_113 Group Title
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    \[\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 -c_1\\ 2y+1=e^{2x+2c_2-c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}(e^{c_3}e^{2x} - 1)\\ y=\frac{1}{2}(Ce^{2x} - 1) \] This looks right for the first question.

    • one year ago
  9. DanielM_113 Group Title
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    I forgot the -1 on the previous one.

    • one year ago
  10. amistre64 Group Title
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    \[ dy/dx = 2y+1 \] is seperable in the sense that we can form \[g(y)~dy = f(x)~dx\] \[ dy/dx = 2y+1 \] \[ \frac{1}{2y+1}dy = dx \] \[\int \frac{1}{2y+1}dy =\int dx \] \[\frac12ln(2y+1) =x+C \] keep in mind the that +C is an all encompassing constant: therefore 2C and c1+c2 and the like are contained within it. \[ln(2y+1) =2x+C \] \[2y+1 =e^{2x+C} \] \[2y =e^{2x+C}-1 \] \[y =\frac12e^{2x+C}-\frac12 \] \[y =\frac12Ce^{2x}-\frac12 \] so tahts fine

    • one year ago
  11. amistre64 Group Title
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    y' + xy = x dy/dx = x(1-y) dy/(1-y) = x dx that can be seperated as well

    • one year ago
  12. DanielM_113 Group Title
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    The first one checks with my answer, thank you! And the second one looks great too.

    • one year ago
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