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Second Order Constant Coefficient Equations

April 5th, 2009 |

Author: Merrin

Constant coefficient equations for short are another class of differential equations with analytic solutions. First we will set our attention to such equations of the form

$a\cfrac{d^{2} y}{dx^{2} } +b\cfrac{dy}{dx} +cy=0$

Compare this equation to another that we already know how to solve, the quadratic equation.

$a\lambda^{2} +b\lambda+c=0$

We can transform the problem of a differential equation into the solution of a quadratic equation if we make the trial substitution of

$y=e^{\lambda x}$

In effect, what this does is make

$\cfrac{d^{2} y}{dx^{2} } \to \lambda^{2} e^{\lambda x} =\lambda^{2} y$
$\cfrac{dy}{dx}\to\lambda e^{\lambda x} =\lambda y\qquad y\to e^{\lambda x} =y$

So the equation to solve

$a\cfrac{d^{2} y}{dx^{2} } +b\cfrac{dy}{dx} +cy=0$

Becomes what is called the characteristic equation

$(a\lambda ^{2} +b\lambda +c)e^{\lambda x} =0$

This doesn’t come out of nowhere. It’s hard to think of any other function besides an exponential which can reduce the problem. We put all instances of y on equal footing and then cancel them. For any finite value of the exponential besides zero, we can just cancel it off and then solve for the parameter lambda using the quadratic equation. Indeed, we see that certain differential equations and algebraic equations are closely linked. In fact, any general equation of the form below can be approached this way.

$a_{1} \cfrac{d^{n} y}{dx^{n} } +a_{2} \cfrac{d^{n-1} y}{dx^{n-1} } +...+a_{n} y=0$

Now back to our problem.

$\lambda_{\pm } =-\cfrac{b}{2a} \pm \cfrac{\sqrt{b^{2} -4ac} }{2a}$

We find that a general solution is of the general form

$y=c_{1} e^{\lambda_{+}}+c_{2} e^{\lambda_{-} }$

From our previous discussion if e^{\lambda x} solves the equation then so does ce^{\lambda x}, so this general solution presumably exhausts all the possibilities. Let us explore some problems and we shall see there still remain a few surprises.

Example 1. Solve the following differential equation

$\cfrac{d^{2} y}{dx^{2} } +3\cfrac{dy}{dx} +2y=0$

Solution 1. We perform the trial substitution

$y=e^{\lambda x}$

We solve the characteristic equation

${\lambda ^{2} +3\lambda +2=0}$
${(\lambda +2)(\lambda +1)=0}$

We can write down the general solution

$y=c_{1} e^{-x} +c_{2} e^{-2x}$

Example 2. Suppose we have found

$\lambda =\pm i$

Which differential equation has this characteristic solution? Also what is the general solution of the differential equation?

Solution 2. We can work backwards

${\lambda ^{2} =-1}$
${\, \lambda ^{2} +1=0}$
${\cfrac{d^{2} y}{dx^{2} } +y=0}$

The solution of the differential equation is the same as before except we will have complex exponentials. We know how to deal with complex exponentials though using Euler’s formula.

$y=c_{1} e^{ix} +c_{2} e^{-ix}$

We can put the solution in a more reasonable form that uses entirely real numbers.

${y=c_{1} (\cos x+i\sin x)+c_{2} (\cos x-i\sin x)}$
${=(c_{1} +c_{2} )\cos x+i(c_{1} -c_{2} )\sin x}$

We can call the first grouped constant A and the second grouped constant including i as B

$y=A\cos x+B\sin x$

Another acceptable form for the general solution is

$y=C\cos (x+\phi )$

I leave it as an exercise to solve for C and phi in terms of A and B.

What the previous solution means is that our function has perpetual oscillations. If this described something physical like a mass hanging on a spring we would call this simple harmonic motion. In real life, this may not be physically realistic because there is always some friction due to air or the material of the spring.

Example 3. What happens if there is a repeated root from the characteristic equation?
Solution 3. Suppose a differential equation has a repeated root. We can construct one if

$b^{2} =4a$

Then

$\cfrac{-b}{2a} \pm \cfrac{\sqrt{b^{2} -4a} }{2a}$

Take the differential equation, if we pick b=2 and a= then c=1.

$\cfrac{d^{2} y}{dx^{2} } +2\cfrac{dy}{dx} +y=0$

The characteristic equation is

${\lambda ^{2} +2\lambda +1=0}$
${(\lambda +1)(\lambda +1)=0}$
$\lambda=-1 \quad \lambda = -1$

Repeating what we did before then the conclusion might be that the general solution is

$y=c_{1} e^{-x} +c_{2} e^{-x}= Ae^{-x} ???$

From a more advanced theory it is known that there must be several linearly independent solutions for a linear differential equation of order two. This basically means the each solution can’t be a combination or proportional to the other solutions. If one solution is known the second can be derived using the following trick.

$y=u(x)e^{-x}$

Now plug it back into the equation and we find that

$\cfrac{d^{2} }{dx^{2} } (ue^{-x} )+2\cfrac{d}{dx} (ue^{-x} )+(ue^{-x} )=0$
$\cfrac{d}{dx} (u'e^{-x} -ue^{-x} )+2(u'e^{-x} -ue^{-x} )+ue^{-x} =0$
${u''e^{-x} -u'e^{-x} -u'e^{-x} +ue^{-x} +2u'e^{-x} -2ue^{-x} +ue^{-x} =0}$
${u''+u'(-2+2)+u(1-2+1)=0}$
${u''=0} \quad {u'=A} \quad {u=Ax+B}$
${y=(B+Ax)e^{-x} }$

We can write the general solution for our repeated root as

$y=c_{1} e^{-x} +c_{2} xe^{-x}$

In general, this is true for any second order linear equation that has a repeated root.

$y=c_{1} e^{-\lambda x} +c_{2} xe^{-\lambda x}$

Example 4. Describe the three possible behaviors for the solution to the following differential equation.

$m\cfrac{d^{2} x}{dt^{2} } +b\cfrac{dx}{dt} +kx=0$

Solution 4. This equation describes the position of a mass attached to a spring which is stretched. The various terms include Hooke’s law F=-kx and a drag force F=-bv. Using Newton’s law

$ma=F_{spring} +F_{drag}$

The properties of the solutions of constant coefficient equations are determined by the characteristic equation. Since this is a physical problem we know that m>0, b>0, and k>0. The coefficients may be renamed

${\cfrac{d^{2} x}{dt^{2} } +2\beta \cfrac{dx}{dt} +\omega _{0}^{2} x=0}$
$2\beta=b/m\quad \omega _{0}^{2}=k/m$

The solution to the characteristic equation is

$\lambda =-\beta \pm \sqrt{\beta^{2} -\omega _{0}^{2} }$

There are three cases where the discriminant is either zero, imaginary, or positive.

If beta<omega_0 the root is complex and the resulting oscillations are damped by a negative real part. The limit beta =0 recovers simple harmonic motion. This type of motion is called underdamped.

$\lambda =-\beta \pm \sqrt{\beta ^{2} -\omega _{0}^{2} } =-\beta \pm i\omega$

$x(t)=Ae^{-\beta t} \cos (\omega t+\phi _{0} )$

If beta =omega _0 the root is real, negative, and repeated, resulting in what is called critically damped motion.

$x(t)=Ae^{-\beta t} +Bte^{-\beta t}$

If beta > omega _0 the roots are real and negative the result is called overdamped motion.

$x(t)=Ae^{-\beta _{1} t} +Be^{-\beta _{2} t}$

Constant coefficient equations occur is basic applications such as harmonic motion and electronic circuits. Their solution is straightforward in terms of a characteristic equation and there are a variety of behaviors of the solutions depending on whether the roots are real, complex, unique, or repeated.

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