SECOND ORDER DIFFERENTIAL EQUATION SOLVERS
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SUNY at Stony Brook MAT 142 Honors Calculus II

Notes on Second Order Differential Equations
Anthony Phillips -- Fall 1998

1. The general second order homogeneous linear differentialequation with constant coefficients looks like

a2y'' + a1y' + a0y =0,

where y is an unknown function of the variable x. If a2 =0 thisbecomes a first order linear equation which we know how to solve. Sowe will consider the case .Then we can divide through by a2 and obtain the equivalent equation

y'' + c1y' + c0y = 0

with c1=a1/a2 and c0=a0/a2. Linear with constant coefficients''means that each term in the equation is a constant times y or aderivative of y. Homogeneous'' excludes equations likey'' + c1y' + c0y =f(x) which can be solved, in certain importantcases, by an extension of the methods we will study here.

2. The method of solution that we will work on involveslooking for an exponential-type solution

Why? Because it works! We substitute in our equation. Thisgives

Since we can divide through and get

If we solve this quadratic equation for ,then willautomatically be a solution of our differential equation. So thesubstitution transforms the differential equation intoan algebraic equation! This equation is called the characteristicequation (associated to the differential equation).

3. The solution of is(quadratic formula'')

In this case a=1, b=c1, c=c0, so the solution is

If the discriminant c12-4c0 is positive, then there are twosolutions, one for the plus sign and one for the minus.

Example 1. If the differential equation is y''-y=0 (so c1=0 and c0=-1) the roots of the characteristicequation are .Both y=ex and y=e-x are solutions. (Check this!)

Example 2. If the differential equation is y''+y'-2y=0 the roots of the characteristicequation are ,so and .Both y=ex and y=e-2x are solutions. (Check this!)

4. If the discriminant c12-4c0 is negative, then the equation has no solutions, unless we enlarge the number field to include ,i.e. unless we work with complex numbers. If c12-4c0=-k2 (we can writeany positive number as k2!) then ik is a square root of c12-4c0since (ik)2 = i2k2 = (-1)k2 = -k2. The solutions of the associatedalgebraic equation are then

Example 3. If we start with the differential equationy''+y=0 (so c1=0 and c0=1) the discriminant is c12-4c0=-4, so 2i is a square root of the discriminant and the solutions of thecharacteristic equation are and .

Example 4. If the differential equation is y''+2y'+2y=0 (so c1=2 and c0=2 and c12-4c0=4-8 =-4). In thiscase the solutions of thecharacteristic equation are ,i.e.and .

5. Going from the solutions of the characteristic equationto the solutions of the differential equation involves interpretingas a function of x when is a complex number.Suppose has real part a and imaginary part ib, so thatwith a and b real numbers. Then

assuming for the moment that complex numbers can be exponentiated so asto satisfy the law of exponents. The factor eax does not causea problem, but what is eibx? Everything will work out if we take

and we will see later that this formula is a necessary consequence ofthe elementary properties of the exponential, sine and cosine functions.

6. Let us try this formula with our examples.

Example 3. For y''+y=0 we found and ,so the solutions are y1=eix and y2=e-ix.The formula gives us and .

Now here is a useful fact about linear differential equations: if y1and y2 are solutions of the homogeneous differential equationy''+c1y'+c2y=0, then so is the linear combination p y1 + q y2 for any numbers p and q. This fact is easy to check (just plug p y1 + q y2 into the equation and regroup terms; note that the coefficients a1 and a0 do not need to be constant for this towork, that p and q can be complex as well as real constants, andthat this works for equations af any order).

Using this fact with the solutions from our example, we notice thatand are both solutions. When we are given a problem with realcoefficients it is customary, and always possible, to exhibit realsolutions. Using the fact about linear combinations again, we can saythat is a solution for any p and q. Thisis the general solution. (It is also correct to call y=p eix +q e-ix the general solution; which one you use depends on the context.)

Example 4. y''+2y'+2y=0. We found and .Using the formula we have

Exactly as before we can take and to get the real solutions and .(Check thatthese functions both satisfy the differential equation!) The generalsolution will be .

7. Repeated roots. Suppose the discriminant is zero:c12-4c0 =0. Then the characteristic equationhas one root. In this case both andare solutions of the differential equation.

Example 5. Consider the equation y''+4y'+4y=0. Herec1 = c0 = 4. The discriminant is .The only root is .Check that both e-2x andx e-2x are solutions. The general solution is theny = p e-2x + q x e-2x.

8. Initial Conditions. For a first-orderdifferential equation the undetermined constant can be adjusted tomake the solution satisfy the initial condition y(0)=y0; in thesame way the p and the q in the general solution of a secondorder differential equation can be adjusted to satisfy initialconditions. Now there are two: we can specify both the value andthe first derivative of the solution for some initial'' valueof x.

Example 5. Suppose that for the differentialequation of Example 2, y''+y'-2y=0, we want a solution withy(0)=1 and y'(0)=-1. The general solution is y=p ex + q e-2x, since the two roots of the characteristicequation are 1 and -2. The method is to write down what theinitial conditions mean in terms of the general solution, and thento solve for p and q. In this case we have

This leads to the set of linear equations p + q = 1, p -2q = -1with solution q = 2/3, p = 1/3. Check that the solutionsatisfies the initialconditions.

Example 6. For the differential equation of Example 4, y''+2y'+2y=0, we found the general solution.To find a solutionsatisfying the initial conditions y(0)=-2 and y'(0)=1 we proceed as in the last example:

So p = -2 and q = -1. Check that the solutionsatisfies the initialconditions.

Tony Phillips
Math Dept, SUNY at Stony Brook

1998-11-30