# Find a differential equation for which the given function is a solution.

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!

• 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 $-2$? Mar 14, 2017 at 22:18
• @Olivier So the solution must be $c_1e^{2x} + c_2 e^{5x}$? where did 3 and -2 come from? Mar 14, 2017 at 22:20

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.
• Do I need to provide initial conditions like $y(x_0) = a, y'(x_0) = b. a,b$ are some constants? Mar 14, 2017 at 22:25
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 $S$.