# F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester's A Modern Introduction to Differential Equations PDF By F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester

ISBN-10: 1852338962

ISBN-13: 9781852338961

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Example text

You should solve by separating variables and then compare your solution to the one we have given. 4 Solutions of the IVP y + xy = 2x; −6 ≤ x ≤ 6, −2 ≤ y ≤ 6 y(0) = −2, 0, 2, 4, and 6 ■ 41 42 CH A P T E R 2: First-Order Differential Equations Rationale Now let’s step back and look at this integrating-factor technique in more generality. Suppose dy + P(x)y = we have written a linear ﬁrst-order differential equation in the standard form dx Q(x). Let’s take P(x), the coefﬁcient of y in the equation, and form the new function μ(x) = 2 e P(x)dx .

Conversely, if y(x) = c is a constant solution, then y = 0, which implies that g(y) = 0 because f (x) = 0 is unlikely in a physical problem. This says that the zeros of g are constant solutions; and, in general, they are the only constant solutions. The preceding analysis can be reﬁned by considering three cases for a separable differential dy = f (x) g(y): (1) g(y) ≡ 1; (2) f (x) ≡ 1; (3) neither (1) nor (2). In case 1, the equation dx equation takes the simple form dy dx dy dx = f (x). If f (x) is continuous on some interval a < x < b, x then the IVP = f (x), y(x0 ) = y0 has a unique solution on (a, b) given by y(x) = y0 + x0 f (r)dr.

Then substitute the original variables for z. Use this technique to solve the equations in Problems 12–14. 12. y − y = 2x − 3 13. (x + 2y)y = 1; y(0) = −1 √ 14. y = 4x + 2y − 1 A homogeneous equation has the form dy/dx = f (x, y), where f (x, y) can be expressed in the form g(y/x) or g(x/y)—that is, as a function of the quotient y/x or the quotient x/y alone. For dy example, by dividing numerator and denominator by x 2 , we can write the equation dx = dy 2−(y/x)2 2x 2 −y 2 3xy in y the form dx = 3(y/x) = g x .