# Method of Undet Coefficient on this DE?

• Mar 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

$\displaystyle y''-5y'+6y=7e^{2x}+cos(2x)$
• Mar 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 $\displaystyle y''-5y'+6=7e^{2x}$ then find Yp for $\displaystyle y''-5y'+6=cos(2x)$ and the two of them together make up the particular solution for the original DE?
• Mar 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

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

Solve the homogeneous DE:

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

Characteristic Equation:

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

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

$\displaystyle m = 2$ or $\displaystyle m = 3$.

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

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

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

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

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

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

Then $\displaystyle y = y_c + y_p$.
• Mar 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?
• Mar 22nd 2010, 10:10 PM
mmattson07
Ok so I set the coefficients equal and found $\displaystyle C1= \frac{1}{52}$ and $\displaystyle C2= \frac{-5}{52}$. I believe I need to find a particular solution for the $\displaystyle 7e^{2x}$ part in which id guess Yp to be $\displaystyle Axe^{2x}$
• Mar 23rd 2010, 12:07 PM
ANDS!
From the homogenous solution to the differential you find that solutions are:

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

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 $\displaystyle x^{k}$, 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:

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

Now substituting into your original d.e.:

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

Terms cancel out leaving:

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

Therefore our particular solution for our exponential is $\displaystyle -7xe^{2x}$
• Mar 23rd 2010, 12:55 PM
mmattson07
Thanks. I calculated the full solution to be

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