Question

y'''+y'+y=sinx

Answer 1

hmm. a non homogeneous linear de huh....i believe this is solveable by undetermined coefficients...are you familiar with that method?

Answer 2

y'''+y''+y'+y=x^2+2x+2

how about this problem?

Answer 3

solveable by undetermined coefficients too

Answer 4

for the first problem

Answer 5

http://www.wolframalpha.com/input/?i=++y%27%27%27%2By%27%27%2By%27%2By%3Dx^2%2B2x%2B2

Answer 6

what do you mean for the first problem?

Answer 7

do u need the solution to the y''+y'+y=sin(x)

Answer 8

yes please

Answer 9

First find the general sol to the homog one

y''+y'+y=0

Do you know how to do that?

Answer 10

quadratic equation

Answer 11

exactly...find the solution to the characteristic equation of it

D^2+D+1=0

Find its roots?

Answer 12

They are complex roots...

Answer 13

Roots are

D^2+D+1=(D+1/2)^2+3/4=0
(D+0.5)^2=-3/4
D1=0.5+i*sqrt(3)/2
D2=0.5-i*sqrt(3)/2

Answer 14

Therefore, the general solution to y''+y'+1=0 is

y(t)=c1e^(-0.5t)sin(sqrt(3)/2t)+c2e^(-0.5t)cos(sqrt(3)/2t)

Answer 15

Sorry, but the roots are

D1=-0.5+i*sqrt(3)/2
D2=-0.5-i*sqrt(3)/2

Answer 16

I dropped the minus sign inform of the 0.5

Answer 17

Are you following?

Answer 18

yes

Answer 19

OK...now find the particular solution

The general one is

y_p(t)=Asin(x)+Bcos(x)

plug y_p(t) in the ODE and solve for A and B

Answer 20

then

Answer 21

y'=Acos(x)-Bsin(x)
y''=-Asin(x)-Bcos(x)

y''+y'+y=sin(x)
-Asin(x)-Bcos(x)+Acos(x)-Bsin(x)+Asin(x)+Bcos(x)=sin(x)
Acos(x)-Bsin(x)=sin(x)

A=0 and B=-1

Answer 22

Therefore, the particular solution is

y_p(t)=-sin(x)

Answer 23

The final general solution is

y(t)= y_h(t) + y_p(t)

y(t)=c1e^(-0.5t)sin(sqrt(3)/2t)+c2e^(-0.5t)cos(sqrt(3)/2t)-sin(x)

Answer 24

got it! thank u.

Answer 25

No problem

Answer 26

@leedomathgeek what problem do you solve. i'm confused.

Answer 27

y''+y'+y=sin(x)