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bayanhorani

i have equation about diff

  • one year ago
  • one year ago

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  1. bayanhorani
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    can help me

    • one year ago
  2. bayanhorani
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    |dw:1350286901878:dw|how to solve this equation step by step

    • one year ago
  3. Woodsmack
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    Hi there. If there is a solution to this equation we have to assume that t=0. THEOREM: If y_1 and y_2 are linearly independent solutions of the homegenous differential equation (a typical notation form) on an interval, and P(x) (leading coefficient) is never 0, the the general solution is given by y(x) = c_1y_1(x) + c_2y_2(x) where c_1 and c_2 are arbitrary constants. We can now say that ay'' + by' + cy =0. Where a, b, and c are constants and a is not equal to 0. I hope that helps. If you are working on the non-homogenous form, as I am now beginning to suspect, then let me know. Don't forget the method of undetermined coefficients.

    • one year ago
  4. Woodsmack
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    \[y(t)=\left( \frac{ t^4 }{ 12 } +c _{1} t + c _{2}\right) e ^{-2t}\]Alright! Finally got it after a mess of simplification. Nice Problem!

    • one year ago
  5. wrwrwr
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    This is an inhomogeneous linear equation, so the general solution may be expressed as a sum of the solution to the complementary equation (with zero on the right) and a particular (whatever) solution to the original equation. Homogeneous part (\(y'' + 4 y' + 4 y = 0\)): -- determine and factor characteristic polynomial: \(p(s) = s^2 + 4 s + 4 = (s + 2)^2\); -- for a real root \(r\), a constant times \(e^{r t}\) is a solution, for a double root the same times \(t\) is also an (independent) solution; -- the equation is of second order, so all complementary solutions are of the form: \(y_h = c_1 e^{-2t} + c_2 t e^{-2t}\) where \(c_1\) and \(c_2\) are some constants. Inhomogeneous part -- we need just a single solution: -- input is something times exponential, so lets guess that a solution is some unknown function times the same exponential: \(y_p = u(t) e^{-2t}\); -- substitute the guess on the left: \(y_p'' + 4 y_p' + 4 y_p = (u'' - 4 u' + 4 u) e^{-2t} + 4 (u' - 2 u) e^{-2t} + 4 u e^{-2t} = u'' e^{-2t}\); -- try to match it with the right: \(u'' e^{-2t} = t^2 e^{-2t}\), exponential is never equal to zero, so we can divide by it: \(u'' = t^2\); -- and integrate twice to get \(u = \frac{t^4}{12}\) (we may ignore constants of integration because we need just a single solution); -- \(y_p = \frac{t^4}{12} e^{-2t}\) is one function that solves your equation. The general solution is sum of the two parts: \(y = y_h + y_p = c_1 e^{-2t} + c_2 t e^{-2t} + \frac{t^4}{12} e^{-2t}\) [exactly as Woodsmack has given above] [In the second part, we could have guessed a polynomial (of degree 4 in this case) instead of using a general function, but this would result in even more calculation. In both cases, the session on undetermined coefficients describes a handy method of organizing those.] This is still more a sketch than a full solution, but maybe it will help you understand where do things come from. If you'd like more details then maybe try Wolfram's Alpha -- it has a nice (and free) step-by-step solver.

    • one year ago
  6. Woodsmack
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    Excellent Reply WR!

    • one year ago
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