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anonymous
 one year ago
check my work ??
anonymous
 one year ago
check my work ??

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0\] 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
 one year ago
Best ResponseYou've already chosen the best response.0if \[y(x)=e^{rx}\] 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: \[r^2+2r+2=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yeah ur right!!! let me try again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0noo wait the polynomial is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think your \[\frac{d^2y}{dx^2}\] is wrong

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3careful you have a **repeated root** !!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh sorry I forgot the constant \[y(x)=Ce^{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the second derivative we need to use the chain rule, or derivative of a product.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx repeated roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah but r is constant so it's derivative would be 0 and u will only get 1 term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooppss!! haha! you are right! r is constant, my bad there!! sorry... take a look at the link @IrishBoy123 posted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is 1 more solution ??

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3the general solution for repeated roots is \( y = (Ax + B ) e^{ax} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok I read the article now so...the solution should be I think \[y(x)=C_{1}e^{x}+C_{2}xe^{x}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3i don't know how you were shown how to to these, but with a differential operator D = d/dx, you take (Da)(Da)y = 0 and let u = (Da)y, so (Da)u = 0, giving \(u = A e^{ax}\) from there \( ( Da)y = A e^{ax},\ \frac{dy}{dx}  ay =A e^{ax} \), solve with integrating factor

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3yes, Ax + B appears in general solution for repeated roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3ok, so now also remember that, with repeated roots, that rule changes :p
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