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Module 1: Practice Problems

Instructions: Work through each problem on paper first, then reveal the solution to check your work. These 10 problems cover classification, verifying solutions, direction fields, existence/uniqueness, and Euler's method.

Problem 1: Classification

Classify each DE by order and linearity:

(a) y'' + 5y' + 6y = 0    (b) (y')² + y = x    (c) y''' - xy = sin x

Solution

(a) Second order, linear. (b) First order, nonlinear (y' is squared). (c) Third order, linear (coefficients depend only on x).

Problem 2: Verify a Solution

Verify that y = 3e2x is a solution to y' - 2y = 0.

Solution

y' = 6e2x. Substituting: y' - 2y = 6e2x - 2(3e2x) = 6e2x - 6e2x = 0. Verified.

Problem 3: Verify a Second-Order Solution

Verify that y = sin 3x is a solution to y'' + 9y = 0.

Solution

y' = 3 cos 3x, y'' = -9 sin 3x. Substituting: y'' + 9y = -9 sin 3x + 9 sin 3x = 0. Verified.

Problem 4: Solve an IVP

Solve dy/dx = 4x³, y(1) = 3.

Solution

Integrate: y = x4 + C. Apply y(1) = 3: 3 = 1 + C, so C = 2. Answer: y = x4 + 2.

Problem 5: Equilibrium Solutions

Find all equilibrium solutions of dy/dx = y² - 5y + 6.

Solution

Set y² - 5y + 6 = 0. Factor: (y - 2)(y - 3) = 0. Equilibria: y = 2 and y = 3.

Problem 6: Direction Field Slopes

For dy/dx = x - 2y, compute the slope at the points (0,0), (1,1), (2,0), and (0,2).

Solution

(0,0): 0 - 0 = 0. (1,1): 1 - 2 = -1. (2,0): 2 - 0 = 2. (0,2): 0 - 4 = -4.

Problem 7: Existence and Uniqueness

Does the IVP dy/dx = y2/3, y(0) = 0 have a unique solution? Justify.

Solution

f(x,y) = y2/3 is continuous. But df/dy = (2/3)y-1/3 is undefined at y = 0. The uniqueness condition fails. Indeed, both y = 0 and y = (2x/3)3/2 satisfy the IVP, so the solution is not unique.

Problem 8: Existence and Uniqueness

Does the IVP dy/dx = x² cos y, y(0) = π have a unique solution near x = 0?

Solution

f = x² cos y is continuous everywhere. df/dy = -x² sin y is continuous everywhere. Both conditions satisfied, so yes, a unique solution exists near (0, π).

Problem 9: Euler's Method

Use Euler's method with h = 0.2 to approximate y(0.4) for dy/dx = 1 + y, y(0) = 0.

Solution

Step 0: f(0,0) = 1. y1 = 0 + 0.2(1) = 0.2, x1 = 0.2.

Step 1: f(0.2, 0.2) = 1.2. y2 = 0.2 + 0.2(1.2) = 0.44, x2 = 0.4.

Answer: y(0.4) ≈ 0.44. (Exact: y = ex - 1, y(0.4) ≈ 0.4918.)

Problem 10: Euler's Method

Use Euler's method with h = 0.1 to approximate y(0.3) for dy/dx = 2y, y(0) = 1.

Solution

Step 0: f = 2(1) = 2. y1 = 1 + 0.1(2) = 1.2, x1 = 0.1.

Step 1: f = 2(1.2) = 2.4. y2 = 1.2 + 0.1(2.4) = 1.44, x2 = 0.2.

Step 2: f = 2(1.44) = 2.88. y3 = 1.44 + 0.1(2.88) = 1.728, x3 = 0.3.

Answer: y(0.3) ≈ 1.728. (Exact: y = e2x, y(0.3) ≈ 1.8221.)

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