# Homework Help: D.E.s Undetermined Coefficients

1. Aug 22, 2010

### Darkmisc

1. The problem statement, all variables and given/known data

y''' + y'' - 6y' = 36x

2. Relevant equations

y'' + y' - 6y =18x^2

3. The attempt at a solution

I've integrated the third order D.E. to get a second order D.E.

Solving the characteristic equation, to find a general solution, I get:

y = c_1exp(-3x) + c_2exp(2x)

However, the answer is given as c_1 + c_2exp(-3x) + c_3exp(2x).

Where is the extra term coming from?

2. Aug 22, 2010

### lanedance

its due to the integration step

if you were just solving the 2nd order DE, the charactristic equation you give would be correct

however as it must only sastisfy the original 3rd order DE you must include the constant. Any constant is a solution of the DE as it becomes zero for all derivatives of y

3. Aug 22, 2010

### gomunkul51

third order ODE have 3 linearly independent parts in the homogeneous solution. :)
y''' + y'' - 6y' = 36x

to find the second equation you replace y(x) with z(x):
y'(x)=z(x)

z'' + z' - 6z = 36x

find the homogeneous answer, the way you regularly do then look at the substitution:
y'(x)=z(x), so y(x) = integral[z(x)], that's where the c_1 term comes from.

also you can solve the cubic equation to get the same answer: r^3 + r^2 - 6r = 0.

if you find the specific answer to the second equation the original specific answer is found, again, through integration.

Good Luck !

4. Aug 22, 2010

### HallsofIvy

I will confess that on my first look at this, I did not notice that it was third order either!

The characteristic equation is $r^3+ r^2- 6r= r(r^2+ r- 6)= r(r- 2)(r+ 3)= 0$.

The characteristic roots are r= 0, r= 2, and r= -3 so the three independent solutions are $e^{0x}= 1$, $e^{2x}$ and $e^{-3x}$.

Now look for a particular solution of the form $Ax+ B$.

Yes, you can integrate the original equation with respect to x first but then you get
$y"+ y'- 6y$$= 18x^2+ C$ where C is the constant of integration and now your characteristic equation is, as you say, $r^2+ r- 6= (r+ 3)(r- 2)$ so that r= -3 and r= 2, giving you you $e^{-3x}$ and $e^{2x}$ as solutions but now you have to try a particular solution of the form $Ax^2+ Bx+ C'$ (and don't confuse this C' with the original C).

Here is yet a third way to solve this problem. Let u= dy/dx and the equation becomes u"+ u'- 6u= 36x. The characteristic equation is $r^2+ r- 6= (r+ 3)(r- 2)= 0$. Trying a particular solution of the form u= Ax+ B, u'= A, and u"= 0 so the equation becomes 0+ A- 6Ax- 6B= 36x. We must have -6A= 36 so A= -6. Then A- 6B= 0 becomes 6B= A= -6 so B= -1.

The solution to u"+ u'- 6u= 36x is $u= y'= C_1e^{3x}+ C_2e^{-2x}- 6x- 1$. Now integrate again to find y: $y(x)= (C_1/3)e^{3x}- (C_2/2)e^{-2x}- 3x^2- x+ C_3$. Of course, since $C_1$ and $C_2$ are arbitrary constants, so are $C_1/3$ and $-C_2/2$ so we could just relabel them as "$C_1$" and "$C_2$".