MATH 2090  Answers to the Review Problems for Exam 1  Fall 2001

1. Solve the initial value problems y′ = 4x\sqrt{y − 1}, y(2) = 1 and y′ = y^{2} sin x, y(0) = 1.

Can you be certain if either of these initial value problems has a unique solution?

Using separation of variables, a solution of y′ = 4x\sqrt{y − 1}, y(2) = 1 is y = 1 + (x^{2} − 4)^{2}. This solution is not unique. For instance, the function y = 1 is also a solution. The existence and uniqueness theorem does not apply to this IVP. Using separation of variables, a solution of y′ = y^{2} sin x, y(0) = 1 is y = sec x. The existence and uniqueness theorem does apply to this IVP, and this solution is unique.

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

(a) y′ + 4y = e^{−x} (c) xy′ − y = 4xy^{2} Hint: make the substitution u = y^{−1}

(b) x + xy^{2} + e^{x^{2}}y \frac{dy}{dx} = 0  (d) y′′ + 10y′ + 25y = 0  (e) y′′ − 2y′ + 5y = 0

(a) y = \frac{1}{3}e^{−x} + Ce^{−4x}  (b) ln(1 + y^{2}) = e^{x^{2}} + C  (c) y = \frac{x}{C − 2x^{2}}

(d) y = c_{1}e^{−5x} + c_{2}xe^{−5x} (e) 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.

(b) Find the general solution of each of the following non-homogeneous differential equations.

(i) y′′ − 4y = 8e^{2x}  (ii) y′′ − 4y = 5 sin x  (iii) y′′ − 4y = 4e^{2x} − 5 sin x

(c) Explain how the results of parts (i) and (ii) above can be used to obtain the answer to part (iii).

(a)  y = 2e^{2x} − e^{−2x}  (b) (i) y = c_{1}e^{2x} + c_{2}e^{−2x} + 2xe^{2x}

(b) (ii) y = c_{1}e^{2x} + c_{2}e^{−2x} − sin(x) (b) (iii) y = c_{1}e^{2x} + c_{2}e^{−2x} + xe^{2x} + sin(x)

(c) The “forcing function” 4e^{2x} − 5 sin x in (iii) is a linear combination of those from (i) and (ii).

4. The function y_{1}(x) = x is solution of the differential equation x^{2}y′′ − 2xy′ + 2y = 0, x > 0. Use the method of reduction of order to find a second linearly independent solution y_{2}(x) on the interval I = (0, ∞). Explain why the solutions y_{1}(x) and y_{2}(x) are linearly independent on I, and find the general solution of this differential equation.

If y = xu, y′ = u + xu′, and y′′ = 2u′ + xu′′. So x^{2}y′′ − 2xy′ + 2y = x^{3}u′′. Since x > 0, the function y = xu is a solution of the DE if u′′ = 0. So u = ax + b for constants a and b, and y = ax^{2} + bx. Since y_{1} = x is a known solution, a “new” solution is y_{2} = x^{2}. These solutions are linearly independent on I (e.g., W [y_{1}, y_{2}] = W [x, x^{2}] = x^{2} ≠ 0 on I), so y = c_{1}x + c_{2}x^{2} is the general solution.

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, and is 50^{∘}F ten minutes later. Assume 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. Then solve it.

\frac{dW}{dt} = k(W − 72), W (0) = 45, W (10) = 50 W (t) = 72 − 27 (\frac{22}{27})^{t/10}

6. Boyle-Mariotte’s law for ideal gases. For a gas at low pressure p (and constant temperature), the rate of change of the volume V = V (p) is equal to −V/p. Determine the volume as a function of p.

If \frac{dV}{dp} = −\frac{V}{p} , then V (p) = \frac{k}{p}, where k is a (positive) constant.