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Then y C1(1 22)e(1 22)x - C1(1 - 22)e(1 - 22)x and so 22 y(0) C1(1 22) - C1(1 - 22) 2C1 22 We conclude that C1 1 2 The general solution to the differential equation is, therefore, and1 y 1 e(1 22)x - 2 e(1 - 22)x 2 This is all fine if the auxiliary equation has distinct real roots.

In particular, an equation of the form F (x, y, y(1), y(2),, y(n)) 0 in which y(k) denotes the kth derivative of y with respect to x, is called an ordinary differential equation of order n.

Examples of differential equations of orders 1, 2, and 3 are y 2 sin x 0 dy dx dx2 2 15.1 Linear Homogeneous Equations 15.2 Nonhomogeneous Equations 15.3 Applications of Second-Order Equations 15.1 3x a dy - 2y 0 dx d3y3 dy 2 b - ex 0 dx If, when f(x) is substituted for y in the differential equation, the resulting equation is an identity for all x in some interval, then f(x) is called a solution of the differential equation.

Thus, f(x) 2 cos x 10 is a solution to y 2 sin x 0 since f(x) 2 sin x -2 sin x 2 sin x 0 for all x.

We call 2 cos x C the general solution of the given equation, since it can be shown that every solution can be written in this form.

In contrast, 2 cos x 10 is called a particular solution of the equation.

Differential equations appeared earlier in this book, principally in three sections.

In Section 3.9, we introduced the technique called separation of variables and used it to solve a wide variety of first-order equations.

In Section 6.5, we solved the differential equation y ky of exponential growth and decay, and in Section 6.6 we studied first-order linear differential equations and some applications.

Thus, if L denotes this operator and if f and g are functions and c is constant, then L(f g) L(f) L(g) L(cf) cL(f) That L has these properties follows readily from the corresponding properties for the derivative operators D, D2,, Dn.

The presence of the exponent 2 on y is enough to spoil the linearity, as you may check.

The theory of nonlinear differential equations is both complicated and fascinating, but best left for more advanced courses.

Chapter 15 Differential Equationstion has the form Second-Order Linear Equations A second-order linear differential equay a1(x)y a2(x)y k(x) In this section, we make two simplifying assumptions: (1) a1(x) and a2(x) are constants, and (2) k(x) is identically zero.

Thus, our initial task is to solve y a1 y a2 y 0 A differential equation for which k(x) 0 is said to be homogeneous.

To solve a first-order equation required one integration and led to a general solution with one arbitrary constant.

By analogy, we might expect that solving a second-order equation to involve two integrations and thus that the general solution would have two arbitrary constants.

In fact, a second-order homogeneous linear differential equation always has two fundamental solutions u1(x) and u2(x), which are independent of each other (i.e., neither function is a constant multiple of the other).

By the linearity of the operator D2 a1 D a2, C1u1(x) C2 u2(x) is also a solution.

To test this possibility, we first write the equation in the operator form (1) Now (D2 a1D a2)erx D2(erx) a1D(erx) a2 erx r2erx a1rerx a2erx erx (r 2 a1r a2) The latter expression is zero, provided (2) r2 a1r a2 0 Equation (2) is called the auxiliary equation for (1) (note the similarity in form).

It is an ordinary quadratic equation and can be solved by factoring or, if necessary, by the Quadratic Formula.

There are three cases to consider, corresponding to whether the auxiliary equation has two distinct real roots, a single repeated root, or two complex conjugate roots.

Theorem A Distinct Real Roots If r1 and r2 are distinct real roots of the auxiliary equation, then the general solution of y a1y a2 y 0 is y C1er1x C2er2x (D2 a1D a2 )y 0 The Auxiliary Equation Because Dx(erx) rerx, it seems likely that erx will EXAMPLE 1 Find the general solution to y 7y 12y 0.

SOLUTION The auxiliary equation r2 7r 12 (r 3)(r 4) 0 has the two roots -3 and -4.

Since e -3x and e -4x are independent solutions, the general solution to the differential equation is y C1e -3x C2e -4x Section 15.1 Linear Homogeneous Equations 777 EXAMPLE 2and y(0) 22.

Find the solution of y - 2y - y 0 that satisfies y(0) 0 SOLUTION The auxiliary equation r2 - 2r - 1 0 is best solved by the Quadratic Formula.

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