Slope fields and Euler’s method, question 12

Topic: Slope fields and Euler’s method
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §11.1Key results

Consider the initial value problem below.

\frac{dy}{dx} = x - 2y, \qquad y(0) = 1

Check that $y = \dfrac{x}{2} - \dfrac{1}{4} + Ce^{-2x}$ solves the equation for every constant $C$ , and choose $C$ . The solution falls at first and then rises. Find its minimum value.

Your answer

An exact expression, or a decimal to three places.

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Theory UBC Math 100 §11.1Key results

Worked solution

Strategy

Verify the family by substitution, let the initial condition fix C, then find where the slope is zero. The equation itself says where that can happen: dy/dx = x − 2y is zero only on the line y = x/2.

Working

$y' = \tfrac{1}{2} - 2Ce^{-2x}, \qquad x - 2y = x - x + \tfrac{1}{2} - 2Ce^{-2x} = \tfrac{1}{2} - 2Ce^{-2x} \quad\checkmark$
Both sides agree for every C.
$y(0) = -\tfrac{1}{4} + C = 1 \implies C = \tfrac{5}{4}, \qquad y = \tfrac{x}{2} - \tfrac{1}{4} + \tfrac{5}{4}e^{-2x}$
The initial condition chooses one solution from the family.
$y' = \tfrac{1}{2} - \tfrac{5}{2}e^{-2x} = 0 \iff e^{-2x} = \tfrac{1}{5} \iff x = \tfrac{\log 5}{2}$
y′ is negative before this x and positive after it, so the solution has its minimum here.
$y = \frac{\log 5}{4} - \frac{1}{4} + \frac{5}{4}\cdot\frac{1}{5} = \frac{\log 5}{4} \approx 0.402$
Substitute into the solution; as it should, y is half of x, on the line where the slope is zero.

Answer

y_{\min} = \frac{\log 5}{4}

Check

Either side of the minimum the solution is higher: y(0.7) ≈ 0.4082 and y(0.9) ≈ 0.4066, against 0.4024 at x = (log 5)/2 ≈ 0.805.

Where this goes wrong

Reporting where the minimum is, x = (log 5)/2, instead of the minimum value. The slope x − 2y is zero only where y = x/2, so the value is half the location.