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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!)

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    $\begingroup$ 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. $\endgroup$ – Rebellos Nov 20 '17 at 14:37
  • $\begingroup$ Are there other restrictions? In this generality, also $y''=45e^{3x}+16e^{2x}$ is a valid second order linear DE. $\endgroup$ – LutzL Nov 20 '17 at 14:42
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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?

P.S. There are more examples of linear second order differential equation satisfied by the given function. See for example LutzL's comment.

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  • $\begingroup$ I'm so sorry, could you please explain me how to find A and B? $\endgroup$ – Rey Nov 20 '17 at 15:00
  • $\begingroup$ 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 $Ax+B$. $\endgroup$ – Robert Z Nov 20 '17 at 15:09

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