# Thread: Method of Undet Coefficient on this DE?

1. ## 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

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

2. 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?

3. 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 $y = y_c + y_p$.

4. 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?

5. 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 $Axe^{2x}$

6. 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 $-7xe^{2x}$

7. Thanks. I calculated the full solution to be

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