Question

nonhomogeneous ODE - Method of Undetermined coefficients question

Answer 1

they give you yh; and ask you to determine yp correct?

Answer 2

How do I make the correct guess? I seem to get it wrong all the time

Answer 3

my teacher taught us a shortcut which involved taking the derivative of the right side thingy

Answer 4

are you guessing from a table?

Answer 5

...but guessing works too...

Answer 6

the wronskian always works, yes

i learned this the long way tho, which builds up to the Wronskian

Answer 7

what's wronskian?

Answer 8

its a fancy name for using a system of determinants

Answer 9

...oh....fancy....

Answer 10

We use Wronskian to check if a set of functions is linearly dependent or independent.

Answer 11

anyway... x^2 - x looks like \[\Large c_1 x e^{0x} + c_2 x^2 e^{0x}\]
doesnt it?

Answer 12

here you are given second degree polynomial so your guess should be
\[\Large y_{p}=Ax^2+Bx+C\]
but since a constant is also there in the complementary solution ,Multiply with x the yp to remove the duplication
hence correct guess is
\[\Large y_{p}=Ax^3+Bx^2+Cx\]

Answer 13

Abel's theorem is a good Alternative for Wronskian.

Answer 14

\[y_p = A(x)+B(x)e^{-x}\]
\[y'_p = A'(x)+B'(x)e^{-x}-B(x)e^{-x};~~A'(x)+B'(x)e^{-x}=0\]
\[y'_p = -B(x)e^{-x}\]
\[y''_p = -B'(x)e^{-x}+B(x)e^{-x}\]

y'p + y''p = x^2-x

-B(x)e^{-x} -B'(x)e^{-x} +B(x)e^{-x}=x^2-x
-B'(x)e^{-x}=x^2-x

\[A'(x)+B'(x)e^{-x}=0\]\[0A'(x)-B'(x)e^{-x}=x^2-x\]

Answer 15

B' = 1(x^2-x) - 0
A' = 0 - e^(-x)(x^2-x)

integrate to determine A and B

Answer 16

\[B = \int B'=\frac{1}{3}x^3-\frac 12x^2\]
\[A = \int A'=\int -x^2e^{-x}dx+\int xe^{-x}dx\]

Answer 17

the guess i found above differentiate it
\[\Large y'=3Ax^2+2Bx+C\]
differentiate again
\[\Large y''=6Ax+B\]
put in the given differential equation values of y' and y'' you will have
\[\Large (6Ax+B)+(3Ax^2+2Bx+C)=x^2-x\]
compare coefficients on both sides
\[\Large 3A=1\]
\[\Large 6A+2B=-1\]
\[\Large C=0\]
solve the above to get values of A and B put in the following
\[\Large y_{p}=Ax^3+Bx^2+Cx\]
this is your yp
and complete solution is sum of both complementary and particular solutions
\[\Large y=y_{c}+y_{p}\]

Answer 18

@amistre64 I kinda with the direction you are going, this is what I got

Answer 19

I have 2B?

Answer 20

you got it correctly . just compare coefficients on both sides.

Answer 21

iron, that is a suitable method if you are taking the "table" route

Answer 22

@sami-21 You mean like \[3Ax ^{2}=x ^{2}\]

Answer 23

yes
so
3A=1
A=1/3

Answer 24

Amistre, I had no idea this route had a name, I am just following a method from notes from a friend and hoping for the best thats its correct..

Answer 25

and the general solution is Yh+Yp?

Answer 26

yes

Answer 27

Hmm, these questions aren't too bad, just gotta make sure I guess right at the beginning.

Answer 28

lol, the "table" method lists suitable guesses for Yp based on the structure of the equating function; in this case it was x^2 - x

Answer 29

yup.

Answer 30

@sami-21 In regards, to your post did you mean a constant in my homogenous solution

Answer 31

yes. there is constant A in homogeneous Part

Answer 32

@sami-21 Ok I got my general solution to be

Answer 33

Just wondering in the 2nd part of the question, the IVP. what equation should I derive?

Answer 34

you cannot use the value of A and B for Homogeneous Part . they should be there
as A and B .
you will find A and B of general solution using Initial conditions.
you general solution should be
\[\Large y=A+Be^{-x}+\frac{1}{3}x^3-\frac{3}{2}x^2+3x\]

Answer 35

I think I am going to end up with simultaneous equations

Answer 36

A+B=1

Answer 37

From the second IVP I got B=-2 and therefore A=-1

Answer 38

you should have
-1=-B+3
B=4

Answer 39

@sami-21 I multiplied everything by -1 to change signs. Can I not do that?

Answer 40

you can do that .
but you should get the same B=4

Answer 41

Algebra fail haha

Answer 42

Silly mistake.