anonymous
  • anonymous
check my work ??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
nah
anonymous
  • anonymous
\[\frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0\] Let \[y(x)=e^{rx}\]\[\implies r^2e^{rx}+2re^{rx}+e^{rx}=0\]\[e^{rx}(r^2+2r+1)=0\] On solving the polynomial, \[r=-1\] therefore, \[y(x)=e^{-x}\]
anonymous
  • anonymous
if \[y(x)=e^{rx}\] then \[\frac{dy}{dx}=re^{rx}\] and \[\frac{d^2y}{dx^2}=e^{rx}+r^2e^{rx}\] so we get: \[r^2e^{rx}+2re^{rx}+2e^{rx}=0\] the polynomial you will get is slightly different: \[r^2+2r+2=0\]

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anonymous
  • anonymous
why is that ??
anonymous
  • anonymous
oh yeah ur right!!! let me try again
anonymous
  • anonymous
noo wait the polynomial is correct
anonymous
  • anonymous
I think your \[\frac{d^2y}{dx^2}\] is wrong
IrishBoy123
  • IrishBoy123
careful you have a **repeated root** !!!
anonymous
  • anonymous
Oh sorry I forgot the constant \[y(x)=Ce^{-x}\]
anonymous
  • anonymous
for the second derivative we need to use the chain rule, or derivative of a product.
IrishBoy123
  • IrishBoy123
http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx repeated roots
anonymous
  • anonymous
yeah but r is constant so it's derivative would be 0 and u will only get 1 term
anonymous
  • anonymous
ooppss!! haha! you are right! r is constant, my bad there!! sorry... take a look at the link @IrishBoy123 posted
anonymous
  • anonymous
Yeh im checking it
anonymous
  • anonymous
There is 1 more solution ??
IrishBoy123
  • IrishBoy123
the general solution for repeated roots is \( y = (Ax + B ) e^{ax} \)
anonymous
  • anonymous
ok I read the article now so...the solution should be I think \[y(x)=C_{1}e^{-x}+C_{2}xe^{-x}\]
IrishBoy123
  • IrishBoy123
i don't know how you were shown how to to these, but with a differential operator D = d/dx, you take (D-a)(D-a)y = 0 and let u = (D-a)y, so (D-a)u = 0, giving \(u = A e^{ax}\) from there \( ( D-a)y = A e^{ax},\ \frac{dy}{dx} - ay =A e^{ax} \), solve with integrating factor
IrishBoy123
  • IrishBoy123
yes, Ax + B appears in general solution for repeated roots
anonymous
  • anonymous
I only knew one method u take the polynomial and with the roots u do \[y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}\]
IrishBoy123
  • IrishBoy123
ok, so now also remember that, with repeated roots, that rule changes :p
anonymous
  • anonymous
kk

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