Question

Need help with an undetermined coefficient problem. Will fan if all steps are shown! Problem inside.

Answer 1

\[y'''+y'=2+sinx\]

Answer 2

homogeneous: y'''+y'=0
k^3+k=0 <--characteristic equation
k(k^2+1)=0
k=0 or k=i or k=-i
y=e^(0t) or y=e^(it) or y=e^(-it)
y=c or y=c e^(it) or y=ce^(-it)
particular:
sin(x)=(ie^(-ix)-ie^(ix))/2
yp=c4 x e^(ix) + c5 x e^(-ix) + c6 x <-- x's are added in so that the particular differs from the homogeneous solutions
solve for c4,c5,c6
then y=yp+c1+c2 e^(it)+c3 e^(-it)

Answer 3

Thanks. Question: I thought for the particular you use the f(x)

Answer 4

you do use the f(x) for the particular except when the f(x) is equal to the homogeneous. in that case, you need to multiply by x; otherwise, the particular will be 0 since the homogeneous is 0.

so for f(x)=2+sin(x)
the first term, 2, needs to be multiplied by x (c6 x) otherwise y'''+y' for y=c*2 will be 0
since y=c is a homogeneous solution
similarly, for the e^(it) component of sin(x), y'''+y' for y=c e^(it) will be 0 since y=c e^(it) is a homogeneous solution
so we need to use y=c*x*e^(it) for the particular, so that y'''+y' is not equal to 0

Answer 5

also on an unrelated note, you can express all the complex exponentials in terms of sin and cosine if you prefer these over imaginary numbers

Answer 6

yeah I did, can I show you what I have done so far?

Answer 7

sure :)

Answer 8

\[y'''+y'=2+sinx\] \[y_c=c_1+c_2cosx+c_3+sinx\]

Answer 9

\[y_p=Asinx+B\]
\[y'_p=Acosx\]
\[y''_p=-Asinx\]
\[y'''_p=-Acosx\]

Answer 10

is this right?

Answer 11

close, however, the particular cannot be equal to the complementary
if y=Asinx, then y'''+y' will equal 0 (since sinx is a complementary solution)
to fix this, multiply by x:
yp=A x sin(x)
B has the same problem
yp=A x sin(x)+Bx

Answer 12

the way you have written it, yp is a subset of yc

Answer 13

Ohhhhhhh

Answer 14

So is there a case where I don't have to tack on an X.

Answer 15

yup, if the the complementary solution did not have a c3 sin(x) term, you could write
yp=Asin(x)+Bx
similarly, if the complementary did not have a c1 term,
yp=Axsin(x)+B would work
and if the complementary had neither a c1 term not a c3 sin(x) term, then
yp=Asin(x)+B would be fine

Answer 16

*nor

Answer 17

But since your y_p is derived from your F(x), aren't you essentially copying it every time?

Answer 18

f(x) is independent of the characteristic equation though
the characteristic equation is what determines the complementary solution
if the complementary solution and F(x) share terms, then you need to multiply these shared terms by x

Answer 19

but yes, you are copying F(x) each time and potentially multiplying terms by x or taking derivatives

Answer 20

Oh I think I get it now...So if r values provide an equation similar to F(x), I must tack on an X to avoid cancellation in future steps, right?

Answer 21

I wish I could give you another star lol...you don't how long I was confused about this lol.

Answer 22

yup :) also, http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx explains the method in detail and provides lots of examples

Answer 23

glad i was able to help :)

Answer 24

yeah my friend was telling me about that site. Thanks again.
You may see me posting a lot tonight though...test tomorrow XD

anyway appreciate it!