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REzai

  • one year ago

Determine the solution y(x) satisfying y'' + 2y' + 2y = 0 satisfying y(1)=2 and y'(1) =1. What technique would I use?

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  1. allank
    • one year ago
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    That's a nice constant-coefficient, homogeneous ordinary differential equation. We can get it's characteristic polynomial: \[r^2+2r+2=0\] Then solve for the polynomial's roots (r1 and r2). The fundamental set of solutions will then be of the form \[y_1(t)= e ^{r_1*t} ; y_2(t) = e^{r_2*t} \] If we get a single repeated root r1, the fundamental solution set will be \[y_1(t)= e ^{r_1*t} ; y_2(t) = t*e^{r_1*t} \] Then we write the general solution as \[y(t)=c_1y_1(t) + c_2y_2(t)\] And finally use the initial conditions to solve for c1 and c2.

  2. surjithayer
    • one year ago
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    \[r=\frac{ -2\pm \sqrt{2^{2}-4*1*2} }{2*1 }=-1 \pm \iota \] solution is \[y=e ^{-x}\left[ c1 \cos x+c2 \sin x \right]\] find c1 and c2 from the initial conditions.

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