The given differential equation is a first order linear differential equation. The general form of such an equation is y′ + p(x)y = g(x). In this case, p(x) = -5/x and g(x) = 0.

Theorem 2.7 (Existence and Uniqueness Theorem for Linear First Order ODEs) states that if p and g are continuous on an interval I that contains the point x0, then there exists a unique solution y = φ(x) that satisfies the given initial value problem on the interval I.

In this case, p(x) is not defined at x = 0, so the theorem applies for any x0 ≠ 0. For any such x0, any value of y0 will yield a unique solution to the initial value problem.

The general solution to the differential equation can be found using an integrating factor. In this case, the integrating factor is e^(∫p(x) dx) = e^(5ln|x|) = |x|^5. Multiplying the differential equation by the integrating factor gives |x|^5 y′ + 5x^4 y = 0, which can be rewritten as d/dx[x^5 y] = 0. Integrating both sides with respect to x gives x^5 y = C, where C is the constant of integration. Thus, the general solution is y = C/x^5.

(a) For the initial condition y(1) = 1, we have 1 = C, so the unique solution is y = 1/x^5 for x ≠ 0.

(b) For the initial condition y(1) = -1, we have -1 = C, so the unique solution is y = -1/x^5 for x ≠ 0.

(c) For the initial condition y(-1) = 0, we have 0 = C, so the unique solution is y = 0 for all x.

(d) For the initial condition y(0) = y0, the solution is not defined because the differential equation is not defined at x = 0.

For the initial condition y(1) = -1, a unique solution can be defined on x 0 because p(x) = -5/x is continuous for x 0. However, the solution is not unique for x ≥ 0 because p(x) is not continuous at x = 0.

Question

The given differential equation is a first order linear differential equation. The general form of such an equation is y′ + p(x)y = g(x). In this case, p(x) = -5/x and g(x) = 0.

Theorem 2.7 (Existence and Uniqueness Theorem for Linear First Order ODEs) states that if p and g are continuous on an interval I that contains the point x0, then there exists a unique solution y = φ(x) that satisfies the given initial value problem on the interval I.

In this case, p(x) is not defined at x = 0, so the theorem applies for any x0 ≠ 0. For any such x0, any value of y0 will yield a unique solution to the initial value problem.

The general solution to the differential equation can be found using an integrating factor. In this case, the integrating factor is e^(∫p(x) dx) = e^(5ln|x|) = |x|^5. Multiplying the differential equation by the integrating factor gives |x|^5 y′ + 5x^4 y = 0, which can be rewritten as d/dx[x^5 y] = 0. Integrating both sides with respect to x gives x^5 y = C, where C is the constant of integration. Thus, the general solution is y = C/x^5.

(a) For the initial condition y(1) = 1, we have 1 = C, so the unique solution is y = 1/x^5 for x ≠ 0.

(b) For the initial condition y(1) = -1, we have -1 = C, so the unique solution is y = -1/x^5 for x ≠ 0.

(c) For the initial condition y(-1) = 0, we have 0 = C, so the unique solution is y = 0 for all x.

(d) For the initial condition y(0) = y0, the solution is not defined because the differential equation is not defined at x = 0.

For the initial condition y(1) = -1, a unique solution can be defined on x > 0 because p(x) = -5/x is continuous for x > 0. However, the solution is not unique for x ≥ 0 because p(x) is not continuous at x = 0.

Knowee AI · Accepted Answer