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anonymous

  • one year ago

Eliminate the arbitrary constants. Any solutions. Problem is given by the picture.

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  1. anonymous
    • one year ago
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  2. Haseeb96
    • one year ago
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    is there any choices?

  3. Haseeb96
    • one year ago
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    Well, there is no any solution to eliminate those constants. Because both are different

  4. UsukiDoll
    • one year ago
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    something's off. it's like the general solution for a second order ode with a repeated root.

  5. UsukiDoll
    • one year ago
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    well I can't stay up... I'm off to bed.

  6. anonymous
    • one year ago
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    the book say the answer is y"-2y'+y=0 i do not know how to solve it.

  7. anonymous
    • one year ago
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    but the problem is to eleminate the arbitrary constants which are c1 and c2.

  8. imqwerty
    • one year ago
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    all u have to do is differentiation :) hope this helps

  9. anonymous
    • one year ago
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    but how? I am having trouble of iy, i know how to get the derivative but i do not know how to eliminate.

  10. freckles
    • one year ago
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    @njwild you keep saying you want to eliminate the arbitrary constants but you said the answer was a differential equation so it makes me think you want to work from y=c1e^x+c2xe^x to get y''-2y'+y=0 . But if you are looking for the constants than you initial conditions.

  11. anonymous
    • one year ago
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    but how to solve it by a given problem to get the answer with solution.

  12. imqwerty
    • one year ago
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    can u find the value of y' and y' by differentiation??

  13. anonymous
    • one year ago
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    after differentiation imqwerty , what is the steps of rliminating or the first to eliminate until last to get the answer

  14. anonymous
    • one year ago
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    @imqwerty I get the y' and y" but i do not know what is the first to eliminate until last to get the answer.

  15. imqwerty
    • one year ago
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    wait m tellin :)

  16. imqwerty
    • one year ago
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    it would take too long to type btw heres the solution - https://www.youtube.com/watch?v=HIdKpnWb2Ws

  17. imqwerty
    • one year ago
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    nt exactly the same but it may help u

  18. freckles
    • one year ago
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    If you are trying to get the differential equation from the solution here is my explanation: \[y=c_1e^x+c_2xe^{x} \\ \text{ the } x \text{ next to the other } e^x \text{ tells us we have a repeated solution \to the } \\ \text{ I will call it the characteristic equation } \\ \text{ since the solution is } y=c_1e^{1 \cdot x}+c_2 xe^{1 \cdot x} \\ \text{ then the solution to the charateristic equation } is r=1 \\ r-1=0 \\ (r-1)^2=0 \text{ since the solution is repeated } \\ r^2-2r+1=0\] from here it should be easy to write the differential equation

  19. freckles
    • one year ago
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    but to figure out what the constants are you need conditions such as y(a)=b and y'(c)=d whatever a,b,c, and d are given as

  20. freckles
    • one year ago
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    other than that I have no idea what you mean by eliminate the arbitrary constants

  21. UsukiDoll
    • one year ago
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    @freckles that's what I had typed earlier because we had to look for the repeated root, but it was getting late and the asker drove me nuts, so I was erasing my posts before I logged off. Users should stop posting generic questions and add more details instead of one liners.

  22. anonymous
    • one year ago
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    differentiate twice and then eliminate c1 and c2 from three equations.

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