Question

y"+y'=3x^2
Can someone explain why, if we use undetermined coefficients on this, we need to guess that Yp has an x^3 in it ? I can't seem to find the rule about this.

Answer 1

Isn't it only sensible that if a sum of two derivatives has an x^2 that the integral would have an x^3? Or am I being too simplistic?

Answer 2

Mr. Math should know.

Answer 3

the problem is the homogenous solution contains a e^0 i.e. a constant, which matches one of the derivative of the right-hand side

Answer 4

Assume you have some polynomial P(x) as as solution of the given differential equation. Now, the given DE indicates that the sum of the second derivative added to the first derivative gives us a second degree polynomial. That means our assumption should be a polynomial at which its first and second derivative contains a term of x raised to power 2. Does that make sense?

Answer 5

In other words, adding the term with x^3 so that we will get x^2 for y'.

Answer 6

if you had e.g.
y'' + y' +y= 3x^2
then the particular solution would be order 2

Answer 7

Yes, if we had y in there. But since y' has the largest degree, then it should be of a degree 2. (y' should have a degree 2).

Answer 8

i explained on the other post Turingtest, please go check it when you see this

Answer 9

im curious that since the derivative of a function IS a function; why we shouldnt be able to play with this as a: y' + y = 3x^2

Answer 10

is that possible? or am i being too niave :)

Answer 11

You can deal with it as \(y'+y=x^3\).

Answer 12

I just integrated both sides, as you can see.

Answer 13

thanks guys, makes sense :D
(I was afk)

Answer 14

@ami
wolfram says we can get to the same answer, by integrating the solution to your problem
http://www.wolframalpha.com/input/?i=y%27%2By%3D+3x%5E2
http://www.wolframalpha.com/input/?i=y%27%27%2By%27%3D+3x%5E2
nice insight!