Question

Anyone here good at differential equations?

Answer 1

\[y''+3y'+2=1+3x+x^2\]

Answer 2

/is it 2y or just 2?

Answer 3

sorry 2y

Answer 4

ok, so find the CF first... by solving m^2+3m+2=0

Answer 5

I could only find the solution for the one in the left

Answer 6

m=-1,-2

Answer 7

yeah

Answer 8

p.I. =(1/(D^2+3D+2)) (1+3x+x^2)

Answer 9

or you may use the method of variation of parameters

Answer 10

but the other one is like this \[(-3+\sqrt{5})/2\] and the other one \[(-3-\sqrt{5})/2\]

Answer 11

what do you mean? ? The other?????

Answer 12

I gotta use undetermined coefficients @jango_IN_DTOWN

Answer 13

yes the other one

Answer 14

/oh, you should have mentioned in the question: solve by the method of undetermined coefficients.

Answer 15

sorry, sorry. I-m worried bout this

Answer 16

I have the answer but I can't get to it because of those strange roots of the other one

Answer 17

So , comparing with the equation y''+Py'+Qy=X, we have P=3,Q=2 and X is a polynomial of degree 2. Both are non-zero. So assume y_p=ax^2+bx+c

Answer 18

then you need to determine a,b,c.

Answer 19

wait a minute

Answer 20

Ill send a pic

Answer 21

I got the P.I as (1/2)x^2

Answer 22

you are correct
however i dont get to it

Answer 23

Ok. here comes the solution

Answer 24

as I said, the pi is of the form ax^2+bx+c where you need to determine a,b,c..

Answer 25

now the p.i. need to satisfy the given differential equation so
(ax^2+bx+c)''+3(ax^2+bx+c)'+2(ax^2+bx+c)=1+3x+x^2
or, 2a+(6ax+3b)+2(ax^2+bx+c)=1+3x+x^2
or, 2ax^2+(2b+6a)x+(2a+3b+2c)=1+3x+x^2
now compare both the sides and you get a=1/2,b=0,c=0
so p.i.= (1/2)x^2

Answer 26

Thats what i have done

Answer 27

Thats what i have done

Answer 28

@jango_IN_DTOWN thats the pic

Answer 29

I dont have any idea what you did. The first part , finding the CF is fine. Rest part you write from the solution I did

Answer 30

I found the roots of the olynomial in the right and then i found the solutions, then I derivated them acording to the function in the left then i added each of them

Answer 31

I'll show you an excercise i got correct so you gonna have and idea of the method i use

Answer 32

Method of undetermined coefficients is the method I used. The method you used is not the method you asked for solution. If you need a table , which have all the possible preferable cases of the method of undetermined coefficients, then I can provide

Answer 33

The second problem is good.. Do you need the table of undetermined coefficients?

Answer 34

No, but i wanna follow the steps i used in that excercise to the one im asking for. Any idea.

Answer 35

which prob you need to see?

Answer 36

@nvafer your problem which got correct was in the format of case 3 a in picture 3 I posted. So you got it correct

Answer 37

In the given problem, it is of the format of a polynomial of degree 2 and P, Q are both non-zero. So it is of the Case 1 a) of pic 1.

Answer 38

Thanks man i know where the fail is i think that was because im tired xD

Answer 39

Ill show you a pic

Answer 40

@jango_IN_DTOWN

Answer 41

\[y''+3y'+2=1+3x+x^2 \] using undetermined coefficients

r^2 + 3r + 2=0
r=-1 , r=-2
in this case with two real roots, the homogeneous solution is form
\[\large y _{h}=C _{1}e^{-x}+C _{2}e^{-2x}\]

Answer 42

@jango_IN_DTOWN

Answer 43

good

Answer 44

yeah looks good, the tricky part is deciding what the trial particular solution will look like for different problems