Originally Posted by

**HallsofIvy** No, it's not anything like Euler's equation. It is a linear equation with constant coefficients. AllenCuz **almost** had it. Normally if the right side of a d.e. has $\displaystyle e^{ax}$, you would try a solution of the form $\displaystyle Ae^{ax}$ but if a is already a solution of the characteristic equation, $\displaystyle e^{ax}$already satisfies the homogeneous equation so you try $\displaystyle Axe^{ax}$.

However, if a is a **double** root of the characteristic equation, both $\displaystyle e{ax}$ and $\displaystyle xe^{ax}$ are solutions to the homogeneous equation so try $\displaystyle y= Ax^2e^{ax}$

Here, with a= 3, you should try a solution of the form $\displaystyle y= Ax^2e^{ax}$.

With $\displaystyle y= Ax^2e^{3x}$, $\displaystyle y'= 3Ax^2e^{3x}+ 2Axe^{ax}= (3Ax^2+ 2Ax)e^{3x}$

$\displaystyle y"= 3(3Ax^2+ 2Ax)e^{3x}+ (6Ax+ 2A)e^{3x}= (9Ax^2+ 12Ax+ 2A)e^{3x}$.

Put those into the d.e.