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Please help me with the following question:

Find a second order differential equation with constant coefficients for which the following function is a solution.

$3e^{2x} - 2e^{5x}$

Theorem:

Let $a_1$, $a_2$ be constants and consider the equation $L(y) = y'' + a_1y' + a_2y = 0$. If $r_1$, $r_2$ are distinct roots of the characteristic polynomial p, where $p(r) = r^2+a_1r+a_2$, then the functions $φ_1$, $φ_2$ are defined by $φ_1(x) = e^{r_1x}$ and $φ_2(x) = e^{r_2x}$ are solutions of $L(y) = 0$.

My attempt:

In this case, since $r_1 = 2$ and $r_2 = 5$, so the characteristic polynomial is $(r-2)(r-5) = r^2 - 7r + 10$. Therefore, the differential equation is $y''-7y' + 10y = 0$.

I am wondering how does the solution contain coefficients $3$ and $-2$? Please help!

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  • $\begingroup$ Your question is unclear. You can verify yourself that $y'' - 7y' +10y = 0$ is indeed a differential equation to which your function is a solution. Now what about $3$ and $-2$? $\endgroup$
    – Olivier
    Mar 14, 2017 at 22:18
  • $\begingroup$ @Olivier So the solution must be $c_1e^{2x} + c_2 e^{5x}$? where did 3 and -2 come from? $\endgroup$
    – socrates
    Mar 14, 2017 at 22:20

2 Answers 2

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For any real numbers $c_1$ and $c_2$, $c_1 e^{2x} + c_2 e^{5x}$ is a solution to the differential equation you found. In particular, $$ 3 e^{2x} -2e^{5x} $$ is a solution.

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  • $\begingroup$ Do I need to provide initial conditions like $y(x_0) = a, y'(x_0) = b. a,b$ are some constants? $\endgroup$
    – socrates
    Mar 14, 2017 at 22:25
  • $\begingroup$ If it is asked in the question, yes. Otherwise no. $\endgroup$
    – Olivier
    Mar 14, 2017 at 22:27
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In general, given a fundamental set of solutions $S = \{f_1(x), f_2(x) \dots f_n \}$ which are $C^n$ and $\mathcal{W}(f_1(x), f_2(x) \dots f_n) \neq 0$ then $$\begin{vmatrix} f_1 & f_2 &\dots &f_n & y \\ f_1' & f_2' & \dots & f_n' & y' \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ f_1^{(n)}& f_2^{(n)} & \dots & f_n^{(n)} & y^{(n)} \end{vmatrix} = 0$$

defines a differential equation whose fundamental set of solutions is $S$.

Best Regards.

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