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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
  13. Mendicant_Bias Group Title
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    Nevermind, answered my own question, I believe.

    • 23 days ago
  14. ganeshie8 Group Title
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    :) `1/2C` is as arbitrary as `C`

    • 23 days ago
  15. ganeshie8 Group Title
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    `e^C` is as arbitrary as `C` is too

    • 23 days ago
  16. ganeshie8 Group Title
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    so yeah we can lump `1/2e^C` by a new name `C`

    • 23 days ago
  17. ganeshie8 Group Title
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    im not sure if thats the question though, as you have deleted it before i completely reading...

    • 23 days ago
  18. Mendicant_Bias Group Title
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    Yeah, I understand that, my mistake was that I was thinking of the root-solving methods involving homogeneous higher-order DE's, and thinking of homogeneous solutions, where y_1 and y_2 are multiplied by a *different* arbitrary constant than the original ones obtained from integration, e.g. I thought that \[\frac {1}{2}e^{2x+C}-\frac{1}{2}\]was itself a single solution to a homogeneous higher-order DE, which it clearly *is not*, so I was multiplying it by c_1 in my mind. Totally different technique. Thanks, nonetheless!

    • 23 days ago
  19. Mendicant_Bias Group Title
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    \[y_h = c_1y_1+ c_2y_2\]That's what I was thinking of.

    • 23 days ago
  20. ganeshie8 Group Title
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    I see..

    • 23 days ago
  21. ganeshie8 Group Title
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    \(y = \frac {1}{2}e^{2x+C}-\frac{1}{2}\) is the general solution and represents all the family of curves that satisfy the DE in question... as you said the arbitrary constant is already there and there wont be any more solutions

    • 23 days ago
  22. ganeshie8 Group Title
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    whle cooking up homogeneous solution for higher order DEs, we take all the linear combinations of all the independent solutions

    • 23 days ago
  23. ganeshie8 Group Title
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    so we get : \[\large y_h = \sum c_iy_i\]

    • 23 days ago
  24. ganeshie8 Group Title
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    \(\large y_i\)'s are the independent solutions to the homogeneous part

    • 23 days ago
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