Consider the ODE: $\,y''-y'-2y = e^{3x}.$

If we set $y_p=A e^{3x},$ then \[ y_p'= 3Ae^{3x}, \qquad y_p''= 9Ae^{3x}. \] So \[ 9Ae^{3x}-3Ae^{3x}-2A e^{3x} = e^{3x}, \] implies $A = \dfrac{1}{4}.$ Thus $\,y_p = \dfrac{1}{4}e^{3x}.$

Therefore $\,y_p = \dfrac{1}{4}e^{3x}\,$ is a particular solution of \[y''-y'-2y = e^{3x}.\]

With this method in mind, it may seem reasonable to guess that a particular solution for \[ y''-y'-2y = e^{2x} \] is $y_p=B e^{2x},\,$ with $B$ a constant.

What do you think? 🤔 You should try it! 📝

With this method in mind, it may seem reasonable to guess that a particular solution for \[ y''-y'-2y = e^{2x} \] is $y_p=B e^{2x},\,$ with $B$ a constant.

However, in this case we cannot choose $\,y_p=Be^{2x}.$ 😥

Substituing $\,y_p=Be^{2x},\,$ $\,y_p'= 2Be^{2x},\,$ $\,y_p''= 4Be^{2x}\,$ into the LHS of the ODE we get

$0=$ $4Be^{2x}-2Be^{2x}-2Be^{2x}= e^{2x}$

To find a particular solution of \[ y''-y'-2y = e^{2x} \] we must to consider $\,y_p = u(x) e^{2x}\,$ where $\,u(x)\,$ is a function to be determined.

Again, using the method of undetermined coefficients we obtain the correct particular solution

$ y_p= \dfrac{1}{3}x e^{2x}. $

Design, Images & Applets
Juan Carlos Ponce Campuzano