Slope fields and Euler’s method, question 5: UBC MATH 100 practice

Question 5 of 15Introductory1 mark · 2 minTheory UBC Math 100 §11.1Key results

Compare the equations $2y = 6$ and $\dfrac{dy}{dt} = 2y$ , where in the second $y$ is an unknown function of $t$ . Which statement about their solutions is correct?

Both are solved by $y = 3$ .
The first is solved by the number $3$ . The second is solved by functions of $t$ , such as $y = e^{2t}$ , and in fact by every $y = Ce^{2t}$ .
The second is solved by $y = 0$ and nothing else, since 0 is the only number equal to twice itself.
The second has no solution until a starting value such as $y(0)$ is given.

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Theory UBC Math 100 §11.1Key results

Worked solution

Strategy

An algebraic equation asks for a number; a differential equation relates a function to its own derivative, so it asks for a function. Test the candidates by substituting them.

Working

$2y = 6 \iff y = 3$
A single number solves the algebraic equation.
$y = Ce^{2t}: \quad \frac{dy}{dt} = 2Ce^{2t} = 2y$
The chain rule brings down the 2, so every y = Ce^2t solves the differential equation: a whole family of functions.
$y = 3: \quad \frac{dy}{dt} = 0 \neq 6 = 2y$
The constant 3 does not solve it. The constant 0 does, as the member C = 0 of the family, but it is not the only solution.
A starting value is not needed for solutions to exist; it picks out one of them. With y(0) = 5, for instance, C = 5.

Answer

2y = 6:\ y = 3; \qquad \frac{dy}{dt} = 2y:\ y = Ce^{2t}

Check

At t = 0, y = e^2t has value 1 and slope 2, and 2y = 2: the two sides agree there, and at every other t too.

Where this goes wrong

Solving a differential equation as if y were a number, by treating dy/dt like an unknown value. Its unknown is a whole function of t, so a solution is checked by differentiating it and comparing both sides for every t.