Ecuaciones diferenciales

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CHAPTER 1 Review Problems
The main objective of this set of review problems is practice in the identification of the different types of first-order differential equations discussed in this chapter.In each of Problems 1-36 we identify the type of the given equation and indicate an appropriate method of solution. 1. If we write the equation in the form y′ − (3 / x ) y = x 2 we see that it islinear with integrating factor ρ = x −3 . The method of Section 1.5 then yields the general solution y = x3(C + ln x). This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6leads to the general solution y = x/(C – ln x). We write this equation in the separable form y′ / y 2 = (2 x − 3) / x 4 . Then separation of variables and integration as in Section 1.4 yields thegeneral solution y = C exp[(1 – x)/x3]. If we write the equation in the form y′ + (2 / x) y = 1/ x 3 we see that it is linear with integrating factor ρ = x 2 . The method of Section 1.5 then yields thegeneral solution y = x–2(C + ln x). If we write the equation in the form y′ + (2 / x) y = 6 x y we see that it is a Bernoulli equation with n = 1/2. The substitution v = y −1/ 2 of Eq. (10) in Section1.6 then yields the general solution y = (x2 + C/x)2. 11. This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6 leads to the general solution y = x / (C – 3 ln x). Wewrite this equation in the separable form y′ / y 2 = 5 x 4 − 4 x. Then separation of variables and integration as in Section 1.4 yields the general solution y = 1 / (C + 2x2 – x5). This is a lineardifferential equation with integrating factor ρ = e3 x . The method of Section 1.5 yields the general solution y = (x3 + C)e-3x. We note that Dy ( e x + y e x y ) = Dx ( e y + x e x y ) = e x y + xy e x y, so the given equation is exact. The method of Example 9 in Section 1.6 yields the implicit general solution ex + ey + ex y = C. 19. We write this equation in the separable form y′ / y 2 = ( 2 − 3x...
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