# Find the second order differential equation with given the solution and appropriate initial conditions

Find the second order differential equation with appropriate initial conditions $y(0)$ and $y'(0)$ and a given solution $$y(x) = 5e^{3x} + 4e^{2x} + 6x + 56.$$

How can I find the linear second order differential equation?

(The constants are not the same as the exercise I have to solve, I just need some help understanding a general way of finding the differential equation starting with a solution that looks like the one I've written up there!)

• Please, before submitting questions take some time to work on $LaTeX$, which is the mathematics language code used in this forum ! Here's a MathJax Tutorial :). Also, what are your thoughts on the given problem ? What does the symbol $\times$ represent on the final part of your expression ? Seems weird to be there in any case. – Rebellos Nov 20 '17 at 14:37
• Are there other restrictions? In this generality, also $y''=45e^{3x}+16e^{2x}$ is a valid second order linear DE. – LutzL Nov 20 '17 at 14:42

Hint. Recall the definition of characteristic equation and note that $(z-3)(z-2)=z^2-5z+6$. Then try the following linear second order differential equation $$y''(x)-5y'(x)+6(y)=Ax+B$$ where $A$ and $B$ are two real numbers to be found. Can you take it from here?
• Since $5e^{3x} + 4e^{2x}$ satisfies the homogeneous equation, it suffices to plug $y(x)=6x+56$ in the LHS and compare the result with the RHS $Ax+B$. – Robert Z Nov 20 '17 at 15:09