Method of Undet Coefficient on this DE?

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March 22nd 2010, 12:07 PM
mmattson07

Method of Undet Coefficient on this DE?

Hey would you use the Method of Undetermined Coefficients (Method of Good Guessing) to find a particular sol'n to this DE

$y''-5y'+6y=7e^{2x}+cos(2x)$
March 22nd 2010, 02:19 PM
mmattson07

And you would have to split the particular solution up because the non homogeneous part is the sum of two terms?

In other words find Yp for $y''-5y'+6=7e^{2x}$ then find Yp for $y''-5y'+6=cos(2x)$ and the two of them together make up the particular solution for the original DE?
March 22nd 2010, 03:29 PM
Prove It

Quote:

Originally Posted by mmattson07

Hey would you use the Method of Undetermined Coefficients (Method of Good Guessing) to find a particular sol'n to this DE

$y''-5y'+6y=7e^{2x}+cos(2x)$

Solve the homogeneous DE:

$y'' - 5y' + 6y = 0$

Characteristic Equation:

$m^2 - 5m + 6 = 0$

$(m - 2)(m - 3) = 0$

$m = 2$ or $m = 3$ .

So $y_c = Ae^{2x} + Be^{3x}$ .

Now, notice that we already have a term of the form $e^{2x}$ in $y_c$ . So all we need to worry about now in the particular solution is $C_1\cos{2x} + C_2\sin{2x}$ .

$y_p = C_1\cos{2x} + C_2\sin{2x}$

$y_p' = 2C_2\cos{2x} - 2C_1\sin{2x}$

$y_p'' = -4C_2\sin{2x} - 4C_1\cos{2x}$ .

Substitute these into your DE and solve for $C_1$ and $C_2$ .

Then $y = y_c + y_p$ .
March 22nd 2010, 07:26 PM
mmattson07

EDIT: I substituted the appropriate Yp's and I got

2C2(sin(2x))+2C1(cos(2x))-10C2(cos(2x))+10C1(sin(2x))=7e^2x + cos(2x).

I'm still unsure how to determine C1 and C2 from this?
March 22nd 2010, 10:10 PM
mmattson07

Ok so I set the coefficients equal and found $C1= \frac{1}{52}$ and $C2= \frac{-5}{52}$ . I believe I need to find a particular solution for the $7e^{2x}$ part in which id guess Yp to be $Axe^{2x}$
March 23rd 2010, 12:07 PM
ANDS!

From the homogenous solution to the differential you find that solutions are:

$y=c_1e^{2x}+c_2e^{3x}$ .

However, when you go to solve for the two particular solutions to the differential, you notice that you already have an exponential to the 2x as a solution of your homogenous. So you need to multiple your original "guess" by $x^{k}$ , where k is the smallest exponent needed to get a unique guess that is not a solution to our homogenous differential.

Therefore I would solve my particular as:

$y_p = Axe^{2x}$
$y'_p = Ae^{2x}+2Axe^{2x} \Rightarrow Ae^{2x}[1+2x]$
$y''_p=2Ae^{2x}[1+2x]+Ae^{2x}[2] \Rightarrow 4Axe^{2x}+4Ae^{2x}$

Now substituting into your original d.e.:

$4Axe^{2x}+4Ae^{2x}-5(Ae^{2x}+2Axe^{2x})+6(Axe^{2x})=7e^{2x} \Rightarrow$
$4Axe^{2x}+4Ae^{2x}-5Ae^{2x}-10Axe^{2x}+6Axe^{2x}=7e^{2x}$

Terms cancel out leaving:

$-Ae^{2x}=7e^{2x}$

Therefore our particular solution for our exponential is $-7xe^{2x}$
March 23rd 2010, 12:55 PM
mmattson07

Thanks. I calculated the full solution to be

$\frac{-45}{4}(e^{2x})+\frac{120}{13}(e^{3x})+\frac{1}{52} (cos(2x))-\frac{5}{52}(sin(2x))-7xe^{2x}$