## anonymous one year ago check my work ??

1. anonymous

nah

2. 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, $y(x)=e^{-x}$

3. 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: $r^2+2r+2=0$

4. anonymous

why is that ??

5. anonymous

oh yeah ur right!!! let me try again

6. anonymous

noo wait the polynomial is correct

7. anonymous

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

8. IrishBoy123

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

9. anonymous

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

10. anonymous

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

11. IrishBoy123
12. anonymous

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

13. anonymous

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

14. anonymous

Yeh im checking it

15. anonymous

There is 1 more solution ??

16. IrishBoy123

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

17. anonymous

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

18. 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

19. IrishBoy123

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

20. anonymous

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

21. IrishBoy123

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

22. anonymous

kk