Question

check my work ??

Answer 1

nah

Answer 2

\[\frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0\]
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,
\[y(x)=e^{-x}\]

Answer 3

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:
\[r^2+2r+2=0\]

Answer 4

why is that ??

Answer 5

oh yeah ur right!!! let me try again

Answer 6

noo wait the polynomial is correct

Answer 7

I think your
\[\frac{d^2y}{dx^2}\]
is wrong

Answer 8

careful

you have a **repeated root** !!!

Answer 9

Oh sorry I forgot the constant
\[y(x)=Ce^{-x}\]

Answer 10

for the second derivative we need to use the chain rule, or derivative of a product.

Answer 11

http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx
repeated roots

Answer 12

yeah but r is constant so it's derivative would be 0 and u will only get 1 term

Answer 13

ooppss!! haha! you are right! r is constant, my bad there!! sorry...
take a look at the link @IrishBoy123 posted

Answer 14

Yeh im checking it

Answer 15

There is 1 more solution ??

Answer 16

the general solution for repeated roots is \( y = (Ax + B ) e^{ax} \)

Answer 17

ok I read the article now
so...the solution should be I think
\[y(x)=C_{1}e^{-x}+C_{2}xe^{-x}\]

Answer 18

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

Answer 19

yes, Ax + B appears in general solution for repeated roots

Answer 20

I only knew one method u take the polynomial and with the roots u do
\[y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}\]

Answer 21

ok, so now also remember that, with repeated roots, that rule changes :p

Answer 22

kk