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

March 9, 2017 | | By admin |

By F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester

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Additional resources for A Modern Introduction to Differential Equations

Sample text

Y (n−1) , y (n) = 0 such that y(t0 ) = y0 , y (t0 ) = y1 , y (t0 ) = y2 , . . , and y (n−1) (t0 ) = yn−1 , where y0 , y1 , . . , yn−1 are arbitrary real constants, is called solving an initial-value problem (IVP). The n speciﬁed values y(t0 ) = y0 , y (t0 ) = y1 , y (t0 ) = y2 , . . , and y (n−1) (t0 ) = yn−1 are called initial conditions. Right now we can’t be sure of the circumstances under which we can solve such an initial-value problem. 8. 9 we will consider IVPs for systems of differential equations.

Either way is acceptable, although solving for x as a function of y is easier algebraically. But even if we don’t ﬁnd an explicit solution, we can plot solution curves for different values of the constant C. 2 we use (from top to bottom) C = −7, −5, −3, 0, 3, 5, and 7. 2 dy 2 3 y3 x : the curves x − y + Implicit solutions of dx = 1+y 2 3 3 =C C = −7, −5, −3, 0, 3, 5, and 7; −4 ≤ x ≤ 4, −4 ≤ y ≤ 4 ■ 31 32 CH A P T E R 2: First-Order Differential Equations The third concern we mentioned earlier is that you may not be able to integrate one or both of the sides after you have separated the variables.

Thus, we have a second-order differential d2 s equation dt 2 = C, initial conditions, and some boundary conditions, and we must solve for the unknown function s(t). Now the basic rules of integral calculus tell us that when we ﬁnd the antiderivative of each side of the differential equation in the last paragraph, we get d2 s dt = dt 2 Cdt = Ct + C1 , 17 18 CH A P T E R 1: Introduction to Differential Equations d2 s dt dt 2 where C1 is a constant of integration. But of this last equation gives us s(t) = = ds dt , so ds dt = Ct + C1 .