Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

REzai

  • 2 years ago

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

  • This Question is Open
  1. allank
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy