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Example text

Iii) and determine y. Verify that the solution obtained in this manner agrees with that of Eq. (35) in the text. 7 in connection with second order linear equations. In each of Problems 36 and 37 use the method of Problem 35 to solve the given differential equation. 36. y − 2y = t 2 e2t 37. y + (1/t)y = 3 cos 2t, t >0 40 Chapter 2. 1 we used a process of direct integration to solve first order linear equations of the form dy = ay + b, (1) dt where a and b are constants. We will now show that this process is actually applicable to a much larger class of equations.

21) can be found by substituting the values of the coefficients into the solution (20), but we will proceed instead to solve Eq. (21) directly. First, rewrite the equation as 1 dv/dt =− . 2 Solutions of Some Differential Equations By integrating both sides we obtain t ln |v − 49| = − + C, 5 and then the general solution of Eq. (21) is (24) v = 49 + ce−t/5 , (25) where c is arbitrary. To determine c, we substitute t = 0 and v = 0 from the initial condition (22) into Eq. (25), with the result that c = −49.

For example, for the pendulum, if the angle θ is small, then sin θ ∼ = θ and Eq. (12) can be approximated by the linear equation g d 2θ + θ = 0. (13) 2 L dt This process of approximating a nonlinear equation by a linear one is called linearization and it is an extremely valuable way to deal with nonlinear equations. 1 An oscillating pendulum. 20 Chapter 1. Introduction by linear equations; to study these phenomena it is essential to deal with nonlinear equations. In an elementary text it is natural to emphasize the simpler and more straightforward parts of the subject.

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