# Solving Constant Differencial Equations

• Jun 15th 2007, 09:10 AM
ThePerfectHacker
Solving Constant Differencial Equations
Somebody asked me if I can teach how to solve equations:
$ay''+by' + cy = 0$ on $-\infty < x <\infty \ \ \ (1)$
Where $a\not =0, b, c$ are any real numbers.

It really is easy!

$ak^2 + bk + c = 0 \ \ \ (2)$
Since $a\not = 0$ this is a quadradic equation.
Let $\Delta = b^2 - 4ac$, this is the discriminant.
1) $\Delta >0$: In this case the equation $(2)$ has two real solutions. Call them $r,s$. Then the solution to $(1)$ is given by $C_1e^{rx}+C_2e^{sx}$. Where $C_1,C_2$ are any real numbers.
2) $\Delta = 0$. In this case the equation $(2)$ has one real solution. Call it $t$. Then the solution to $(1)$ is given by $C_1e^{tx}+C_2xe^{tx}$.
3) $\Delta < 0$. In this case the equation $(2)$ has two complex solutions. Since $a,b,c$ are real, the polynomial is a polynomial in real coefficients and hence the complex solutions come as complex conjuages. Thus let $a\pm i\beta$ to the two solutions. Then $C_1 e^{\alpha x}\sin \beta x+C_2 e^{\alpha x}\cos \beta x$ is the solution to $(1)$.