高等数学【线性代数】英文版课件2

(n−1)
(x0 ) = yn−1 ,ຫໍສະໝຸດ Baidu
where y0 , y1 , . . . , yn−1 are constants, is called an initial-value problem.
Ordinary Differential Equations Lecture Notes
Example (1.4.2) Solve the initial-value problem y = e−x y(0) = 1, y (0) = 4. (1.4.1) (1.4.2)
Solution From Example 1.3.7, the general solution to Equation (1.4.1) is y(x) = e−x + C1 x + C2 , where C1 , C2 are arbitrary constants. We now impose the auxiliary conditions (1.4.2).
Ordinary Differential Equations Lecture Notes
Example (1.5.4)
dy Solve (1 + y2 ) dx = x cos x.
Solution. This differential equation is separable: (1 + y2 ) dy = x cos x dx. Integrating both sides of the differential equation yields (1 + y2 ) dy = x cos x dx + C.
1.4. Initial-Value Problem
The unique specification of an applied problem requires more than just a differential equation. Of particular interest is the case of the initial-value problem defined for an nth-order differential equation. Definition (1.4.1) An nth-order differential equation together with n auxiliary conditions of the form y(x0 ) = y0 y (x0 ) = y1 ... y
Ordinary Differential Equations Lecture Notes
So these two solutions must coincide. Therefore, f (x) = A cos ωx + B sin ωx ω f (0) sin ωx = f (0) cos ωx + ω = C1 cos ωx + C2 sin ωx.
Ordinary Differential Equations Lecture Notes
1.5. Separable Differential Equations
We provide some solution techniques for determining the exact solution to certain types of differential equations. The simplest differential equations for which a solution technique can be obtained are the so-called separable equations. Definition (1.5.1) A first-order differential equation is called separable if it can be written in the form dy (1.5.1) p(y) = q(x) dx (that is, if we can separate p(y) dy and q(x) dx)
Ordinary Differential Equations Lecture Notes
Example (1.4.4) The general solution to the differential equation y + ω 2 y = 0, −∞ < x < ∞ where ω is a nonzero constant, is y(x) = C1 cos ωx + C2 sin ωx, where C1 , C2 are arbitrary constants. Solution According to Definition 1.3.5 of the general solution, we have to show the following: (1.4.4) (1.4.3)
Ordinary Differential Equations Lecture Notes
We accept the following important theorem without proof. Theorem (1.4.3) Let a1 , a2 , . . . , an , F be functions that are continuous on an interval I. Then, for any x0 ∈ I, the initial-value problem y(n) + a1 (x)y(n−1) + · · · + an−1 (x)y + an (x)y = F(x), y(x0 ) = y0 , y (x0 ) = y1 , . . . , y(n−1) (x0 ) = yn−1 has a unique solution on I.
has a unique solution. Notice that y = f (x) satisfies this problem, and so y = f (x) is precisely that unique solution to the initial-value problem (1.4.5).
the left-hand side of (1.5.1) can be rewritten as follows p(y) dy d = dx dy p(y) dy dy d = dx dx p(y) dy .
Now integrating both sides of Equation (1.5.3) w.r.t. x yields Equation (1.5.2).
Ordinary Differential Equations Lecture Notes
Proof. We use the chain rule for derivatives to rewrite Equation (1.5.1) in the equivalent form d p(y) dy = q(x). (1.5.3) dx Indeed, since p(y) = d dy p(y) dy ,
Ordinary Differential Equations Lecture Notes
The solution technique for a separable differential equation is given in the following result. Theorem (1.5.2) If p(y) and q(x) are continuous, then Equation (1.5.1) has the general solution p(y) dy = q(x) dx + C, (1.5.2) where C is an arbitrary constant.
Ordinary Differential Equations Lecture Notes
Remark (1.5.3) In differential form Equation (1.5.1), as noted above, can be written as p(y) dy = q(x) dx, and the general solution (1.5.2) is obtained by integrating the left-hand side w.r.t. y and the right-hand side w.r.t. x. This is the general procedure for solving separable equations.
Ordinary Differential Equations Lecture Notes
1
y(x) = C1 cos ωx + C2 sin ωx is a solution to the differential equation (1.4.3) on (−∞, ∞). every solution to (1.4.3) is of the form (1.4.4).
Ordinary Differential Equations Lecture Notes
Setting x = 0 in y(x) = e−x + C1 x + C2 , we see that y(0) = 1 ⇐⇒ 1 = 1 + C2 ⇐⇒ C2 = 0, that is y(x) = e−x + C1 x. Furthermore, differentiating the result yields y (x) = −e−x + C1 . Consequently, y (0) = 4 ⇐⇒ 4 = −1 + C1 ⇐⇒ C1 = 5. Hence y(x) = e−x + 5x is the unique solution to the initial-value problem.
Since f (x) is an arbitrary solution to the differential equation (1.4.3), we can conclude that every solution to (1.4.3) is of the form (1.4.4), that is y(x) = C1 cos ωx + C2 sin ωx and therefore this is the general solution on (−∞, ∞).
2
The first claim is easy (!) to verify. For the second claim, take an arbitrary solution y = f (x) of (1.4.3). Let f (0) = A, f (0) = B. By Theorem 1.4.3, the initial-value problem y + ω2y = 0 y(0) = A, y (0) = B (1.4.5)
Ordinary Differential Equations Lecture Notes
Consider the function ˜(x) = A cos ωx + y B sin ωx. ω (1.4.6)
This function is of the form (1.4.4), and so, by the first claim, is a solution to equation (1.4.3). We notice that evaluations of the function (1.4.6) and its derivative at x = 0 are as follows ˜(0) = f (0) = A, ˜ (0) = f (0) = B, y y which shows that ˜(x) is also a solution to the initial-value problem y (1.4.5).