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Gina

  • 5 years ago

find the differentiation equation,,,,,y'=10^x+y

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  1. anonymous
    • 5 years ago
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    Rewrite as dy/dx - y = 10^x This is in linear first order form, so you see that p(x) is -1 and the integrating factor is e^-x. Then you have to solve the integral 10^x*e^x which you can solve easily by manipulating the exponent.

  2. Gina
    • 5 years ago
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    then what next?

  3. anonymous
    • 5 years ago
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    Well, if you don't know what to do after solving the integral on both sides I don't know what to tell you... It's just isolation after that point

  4. anonymous
    • 5 years ago
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    I'm assuming you know the process on how to solve linear first orders btw.

  5. Gina
    • 5 years ago
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    while ,i just know the basic

  6. anonymous
    • 5 years ago
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    Well do you understand how I arrived at the integral?

  7. Gina
    • 5 years ago
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    not much,sir

  8. anonymous
    • 5 years ago
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    linear first order are d.e in the form dy/dx+ p(x)y=q(x) So first put into that form which I did. Then you need to find something called an integrating factor which is always I=e^(integral of p(x)). So the integreating factor is e^-x. Then you multiple the integrating factor on both sides of the equation. Now for times sake, the left side with the differential will always just be derivative of (y multiplied by the integrating factor). If you foil normally, it is just the product rule so it can be simplified into a derivative. Then multiplying the integrating on the right side now, will yield above. So you have a derivative on the left side, therefore you need to integrate both sides to get rid of it, hence integrate of 10^x*e^-x.

  9. Gina
    • 5 years ago
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    thanks can you help physics?

  10. Gina
    • 5 years ago
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    to the plate plane air capacitor applied potential difference U = 500V in the area of ​​the plates with a = 200 cm square, the distance between d1 = 1,5 mm. plates moved apart to a distance d2 = 15mm. find the energy E1 and E2 of the capacitor before and after razdvizheniya plates, if the source of EMF to razdvizheniem: 1) switched off, and 2) does not turn off. (1) E1 = 14,8 mJ, E2 = 148mkdzh; 2) E1 = 14,8 mJ: E2 = 1,48 mJ

  11. Gina
    • 5 years ago
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    this is the question

  12. anonymous
    • 5 years ago
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    no sorry i took electromag last semester and forgot all about the formulas and most things

  13. Gina
    • 5 years ago
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    while do konw any body who can solve it ,i have to take it 2morrow

  14. anonymous
    • 5 years ago
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    I don't know why you'd use an integrating factor with this. Method of undetermined coefficients is much easier.

  15. anonymous
    • 5 years ago
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    It's not very hard with the integrating factor, and how do you use undetermined coefficients isnt that for second order?

  16. anonymous
    • 5 years ago
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    No, you can use undetermined coefficients for first order. You solve the related homogeneous equation, and 'guess' a particular solution \[y _{p}=Ax+B\], plug that into the original equation, equate coefficients, and then your solution is the solution to the homogeneous plus \[y _{p}\], where A and B are determined by plugging in.

  17. anonymous
    • 5 years ago
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    Interesting, from what I learned they dont teach us to solve first order using undetermined coefficients nor was anyone really a fan of it because it took awhile. I'm guessing any question that can be solved this way can also been done linearly or separably, correct? Anyways, do you mind showing the exact steps on how you would do it that way? I had a few questions with your stuff above and I figure it'd just be faster if you showed the steps.

  18. Gina
    • 5 years ago
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    show us yo steps

  19. anonymous
    • 5 years ago
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    i don't think its any faster but we shall see

  20. Gina
    • 5 years ago
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    ok

  21. anonymous
    • 5 years ago
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    Ugh. I scanned my written work and I am having issues with it. I will get it up asap.

  22. anonymous
    • 5 years ago
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    http://i181.photobucket.com/albums/x277/carlylemiii/Screenshot2011-03-16at61958PM.png THERE! :D Okay, two comments: I always use undetermined coefficients where possible. Unnecessary integration doesn't please me. Second comment, along those same lines—using an integrating factor with this then requires the use of integration by parts. Barf. Just kidding, I have a third comment: I apologize for my switching between x's and t's…I am so used to using t's that I inadvertantly kept changing from x to t. Don't let let confuse you, obviously. Other than that, here you are!

  23. anonymous
    • 5 years ago
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    The original question is 10^x not 10x, that would be why things don't make sense lol. With 10^x the integral is relatively simple so I was confused as to how this could be faster

  24. anonymous
    • 5 years ago
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    But regardless, a few questions. What is the related homogeneous equation form for first orders and how do you solve for Yh? And personally, I find undetermined coefficients algebraically messy, especially compared to variation of paramters for 2nd orders but to each their own I guess.

  25. anonymous
    • 5 years ago
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    My bad. D: Sorrryy. In the case of that, you would have to use an integrating factor. (It was my own bad reading 'cause you did indeed use a carrot, but it still probably would have helped if you put it in the equation editor—probably wouldn't missed it then. Anyway, given a linear first-order ODE, dy/dt=a(t)y+b(t), the related homogeneous is dy/dt=a(t)y. To solve for \[y_{h}\], it's just simple integration; I didn't figure I needed to show separation of variables and such with such a basic diff eq.

  26. anonymous
    • 5 years ago
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    This site needs an edit button: I forgot a closing parenthesis up there. :(

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