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}.

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