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DanielM_113

  • one year ago

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

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  1. DanielM_113
    • one year ago
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    \[\int \frac{dy}{2y+1}=\int dx \\ \ 2 \ln y = x \]

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

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

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

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

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

  7. DanielM_113
    • one year ago
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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.

  8. DanielM_113
    • one year ago
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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.

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

  10. amistre64
    • one year ago
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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

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

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

  13. Mendicant_Bias
    • one month ago
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    Nevermind, answered my own question, I believe.

  14. ganeshie8
    • one month ago
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    :) `1/2C` is as arbitrary as `C`

  15. ganeshie8
    • one month ago
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    `e^C` is as arbitrary as `C` is too

  16. ganeshie8
    • one month ago
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    so yeah we can lump `1/2e^C` by a new name `C`

  17. ganeshie8
    • one month ago
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    im not sure if thats the question though, as you have deleted it before i completely reading...

  18. Mendicant_Bias
    • one month ago
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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!

  19. Mendicant_Bias
    • one month ago
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    \[y_h = c_1y_1+ c_2y_2\]That's what I was thinking of.

  20. ganeshie8
    • one month ago
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    I see..

  21. ganeshie8
    • one month ago
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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

  22. ganeshie8
    • one month ago
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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. ganeshie8
    • one month ago
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    so we get : \[\large y_h = \sum c_iy_i\]

  24. ganeshie8
    • one month ago
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    \(\large y_i\)'s are the independent solutions to the homogeneous part

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