Fall 2000  Math 2090  Exam 1

1. (a) Solve the initial value problem \frac{dy}{dx} = y^{2} sin x, y(0) = 1.

y = sec x solves the IVP

(b) Does the initial value problem in part (a) have a unique solution? Explain. It does. Use the existence and uniqueness theorem for first-order IVPs.

2. Find the general solution of each of the following differential equations.

(a) y′ + 4y = e^{−x}  y = \frac{1}{3}e^{−x} + Ce^{−4x}

(b) y′′ + 10y′ + 25y = 0  y = c_{1}e^{−5x} + c_{2}xe^{−5x}
(c) y′′ − 2y′ + 5y = 0  y = c_{1}e^{x} cos 2x + c_{2}e^{x} sin 2x

3. (a) Solve the initial value problem y′′ − 4y = 0, y(0) = 1, y′(0) = 6.

y = 2e^{2x} − e^{−2x}

(b) Find a particular solution of the differential equation y′′ − 4y = 8e^{2x} + 5 sin x.

y = 2xe^{2x} − sin x

4. Consider the differential equation x^{2}y′′ − 2xy′ + 2y = 0, x > 0.

(a) One solution of this differential equation is y_{1}(x) = x. Use the method of reduction of order to find a second linearly independent solution y_{2}(x) on the interval I = (0, ∞). Briefly explain why the solutions y_{1}(x) and y_{2}(x) are linearly independent on I.

y = Ax^{2} + Bx is a solution by reduction of order, so y_{2} = x^{2} is a second solution

y_{1} = x and y_{2} = x^{2} are not proportional on I, so are linearly independent on I

This can also be established using the Wronskian: W [x, x^{2}] = x^{2} ≠ 0 for x > 0

(b) Find the general solution of this differential equation.

y = c_{1}x + c_{2}x^{2}

5. Newton’s law of cooling states that the rate of change of the temperature of an object with respect to time t is proportional to difference between the temperature, W (t), of the object at time t and the temperature, R, of the surrounding medium.

Suppose that the temperature of the water in my cup is 45^{∘}F at the beginning of class (at time t = 0), and that room temperature R = 72^{∘}F is constant. To determine the temperature of my water at the end of class using Newton’s law of cooling, you would have to solve a certain initial value problem. Write this initial value problem down.

\frac{dW}{dt} = k(W − 72), W (0) = 45