7.10 Constructing differential equations — Practise

Work through these on your own. For each context, write the rate, the sign and the proportional constant before you write the equation. Then check the equation against the description in words.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A population grows at a rate proportional to its size. Show that $P = P_0e^{kt}$ satisfies the differential equation you construct.
A radioactive mass decays at a rate proportional to the mass. Construct the differential equation and show that $m = m_0e^{-kt}$ satisfies it.
A body cools in a room at $15^\circ$ C. Its temperature falls at a rate proportional to the difference from $15$ . Construct the differential equation and show that $T = 15 + Ce^{-kt}$ satisfies it.
A car has constant acceleration $a$ . Construct the differential equation for its velocity $v$ and show that $v = at + C$ satisfies it.
A liquid cools towards a room temperature of $18^\circ$ C at a rate proportional to its excess over $18$ . Construct the differential equation and show that $T = 18 + Ce^{-kt}$ satisfies it.
The price $P$ of an item falls at a rate proportional to the amount by which it exceeds $10$ . Construct the differential equation and show that $P = 10 + Ce^{-kt}$ satisfies it.
The curve $y = Ae^{2x}$ contains a constant $A$ . Show that $y$ satisfies the differential equation $\dfrac{dy}{dx} = 2y$ .
The curve $y = x^3 + C$ contains a constant $C$ . Form a differential equation that does not contain $C$ .
The curve $y = A\sin 2x$ contains a constant $A$ . Show that $y$ satisfies the differential equation $\dfrac{d^2y}{dx^2} = -4y$ .
The curve $y = Ax^2$ contains a constant $A$ . Show that $x\dfrac{dy}{dx} = 2y$ .
The curve $y = Ae^{-x}$ contains a constant $A$ . Show that $y$ satisfies $\dfrac{dy}{dx} = -y$ .
A population $P$ grows at a rate proportional to $(1000 - P)$ . Construct the differential equation and show that $P = 1000 - Ce^{-kt}$ satisfies it.

Answers and concise solutions

1. The differential equation

Answer: The rate is proportional to $P$ and the population grows, so $\dfrac{dP}{dt} = kP$ with $k > 0$ . Substituting $P = P_0e^{kt}$ gives $\dfrac{dP}{dt} = kP_0e^{kt} = kP$ , so the solution satisfies the equation.

2. The differential equation

Answer: The mass falls, so $\dfrac{dm}{dt} = -km$ with $k > 0$ . Substituting $m = m_0e^{-kt}$ gives $\dfrac{dm}{dt} = -km_0e^{-kt} = -km$ , so the solution satisfies the equation.

3. The differential equation

Answer: The temperature difference is $T - 15$ and the body cools, so $\dfrac{dT}{dt} = -k(T - 15)$ with $k > 0$ . Substituting $T = 15 + Ce^{-kt}$ gives $\dfrac{dT}{dt} = -kCe^{-kt} = -k(T - 15)$ .

4. The differential equation

Answer: Acceleration is the rate of change of velocity, so $\dfrac{dv}{dt} = a$ . Substituting $v = at + C$ gives $\dfrac{dv}{dt} = a$ , so the solution satisfies the equation.

5. The differential equation

Answer: The excess is $T - 18$ , and the liquid cools, so $\dfrac{dT}{dt} = -k(T - 18)$ with $k > 0$ . Substituting $T = 18 + Ce^{-kt}$ gives $\dfrac{dT}{dt} = -kCe^{-kt} = -k(T - 18)$ .

6. The differential equation

Answer: The excess is $P - 10$ , and the price falls, so $\dfrac{dP}{dt} = -k(P - 10)$ with $k > 0$ . Substituting $P = 10 + Ce^{-kt}$ gives $\dfrac{dP}{dt} = -kCe^{-kt} = -k(P - 10)$ .

7. The verification

Answer: Differentiate: $\dfrac{dy}{dx} = 2Ae^{2x} = 2y$ . The constant $A$ cancels, so every curve in the family satisfies the same equation.

8. The derivative

Answer: Differentiating once removes the constant: $\dfrac{dy}{dx} = 3x^2$ . Every curve of the family has the same gradient at a given $x$ .

9. The verification

Answer: The first derivative is $2A\cos 2x$ and the second is $-4A\sin 2x$ . Since $A\sin 2x = y$ , the second derivative is $-4y$ , which rearranges to the stated equation.

10. The verification

Answer: $\dfrac{dy}{dx} = 2Ax$ , so $x\dfrac{dy}{dx} = 2Ax^2 = 2y$ . The constant $A$ cancels on both sides.

11. The verification

Answer: $\dfrac{dy}{dx} = -Ae^{-x} = -y$ . The curve falls at a rate equal to its own height.

12. The differential equation

Answer: The rate is proportional to the remaining room, $1000 - P$ , so $\dfrac{dP}{dt} = k(1000 - P)$ with $k > 0$ . Substituting $P = 1000 - Ce^{-kt}$ gives $\dfrac{dP}{dt} = kCe^{-kt} = k(1000 - P)$ .