# Differential Equations, Part I

Suppose ${y}$ is a twice-differentiable real-valued function defined on some open subset of ${\mathbb R}$ that satisfies the differential equation

$\displaystyle y'' + by' + cy = 0,$

where ${b,c\in\mathbb R}$. To find an explicit equation for ${y}$, we can invoke the following trick: suppose ${y=e^{rt}}$ for some ${r\in\mathbb C}$. Then

$\displaystyle y'=re^{rt},\quad y''=r^2e^{rt},$

and so

$\displaystyle r^2e^{rt} + bre^{rt} + ce^{rt} = 0.$

Since the exponential function is never zero, we can divide by ${e^{rt}}$ and obtain the characteristic equation

$\displaystyle r^2 + br + c = 0.$

Now, we will use the following lemma:

Lemma 1 If ${a,b,c\in\mathbb R}$, ${a\ne0}$ then the equation ${ax^2 + bx + c=0}$ has the two complex solutions

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

Proof: Dividing both sides by ${a}$, we have ${x^2+\frac ba x+\frac ca=0}$. Now, as

$\displaystyle \left(x+\frac b{2a} \right)^2 = x^2 + \frac ba x + \frac{b^2}{4a^2},$

we have

$\displaystyle x^2 + \frac ba x + \frac ca = \left(x + \frac b{2a} \right)^2 - \frac {b^2}{4a^2} +\frac ca = 0,$

so that

$\displaystyle \left(x+\frac b{2a} \right)^2 = \frac {b^2}{4a^2}-\frac ca = \frac{b^2-4ac}{4a^2}.$

Taking the square root of both sides, we obtain

$\displaystyle x+\frac b{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a},$

and hence

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a},$

the familiar “quadratic formula.” $\Box$

From this lemma, we have that

$\displaystyle r = \frac{-b\pm\sqrt{b^2-4c}}2.$

For an example, consider the differential equation ${y'' + 3y' + 2y = 0}$. Then the characteristic equation is

$\displaystyle r^2+3r+2r=0,$

which has solutions ${r=1}$, ${r=2}$. So both ${y_1(t) = e^{-t}}$ and ${y_2(t)=e^{-2t}}$ are solutions, as

$\displaystyle y_1''(t) + 3y_1'(t) + 2y_1(t) = e^{-t} - 3e^{-t} + 2e^{-t} = 0,$

and

$\displaystyle y_2''(t) + 3y_2'(t) + 2y_2(t) = 4e^{-2t} - 6e^{-2t} + 2e^{-2t} = 0.$

As differentiation is a linear operator, ${c_1y_1 + c_2y_2}$ are solutions of the ODE for any ${c_1,c_2\in\mathbb R}$.