Separable First order differential equations: notes and practice questions

Consider the following differential equation:

$\left( 2x^{3} + 5 \right)\frac{dy}{dx} = 6x^{2}y^{2}$

$a$ Take $y( - 1) = - 1$, and use Euler's method with a step size of 0.4 to estimate $y$ at $x = 1$.

$b$ By solving the differential equation analytically, calculate the absolute error in your approximation from part (a) at $x = 1.$

The acceleration, $a \text{ ms}^{-2}$, of a particle moving in a straight line at time $t$ seconds, $t \ge 0$, is given by $a = -2v - 8$, where $v \text{ ms}^{-1}$ is the particle's velocity. At $t=0$, the particle is at the origin O and has an initial velocity $v_0 \text{ ms}^{-1}$, where $v_0 > 0$.

By solving an appropriate differential equation, show that the particle's velocity at time $t$ is given by $v(t) = (v_0 + 4)e^{-2t} - 4$.

The particle moves in the positive direction until it reaches its maximum displacement from O at time $T$. Show that $e^{2T} = \frac{v_0+4}{4}$.

Find an expression for the maximum displacement, $s_{\text{max}}$, in terms of $v_0$.

Let $v(T-k)$ represent the particle's velocity $k$ seconds before it reaches $s_{\text{max}}$, where $0 < k < T$. By using the result from part (b)(i), show that $v(T-k) = 4(e^{2k} - 1)$.

Similarly, let $v(T+k)$ represent the particle's velocity $k$ seconds after it reaches $s_{\text{max}}$. Deduce a similar expression for $v(T+k)$ in terms of $k$.

Hence, show that the speed of the particle $k$ seconds before it reaches $s_{\text{max}}$ is greater than or equal to its speed $k$ seconds after it reaches $s_{\text{max}}$.

The rate of change of a certain quantity $R$ with respect to a variable $x$ is given by $R'(x)=\frac{1}{x(M-x)}$, $x \in \mathbb{R}$, $x \neq 0$, $x \neq M$ where $M$ is a positive constant.

The expression for $R'(x)$ can be written in the form $\frac{A}{x} + \frac{B}{M-x}$, where $A, B \in \mathbb{R}$.

Find $A$ and $B$ in terms of $M$.

Hence, find an expression for $R(x)$.

The concentration of a certain chemical product, $C$ (in mol/L), in a reaction vessel at time $t$ (in minutes) can be modelled by the differential equation $\frac{dC}{dt} = \frac{C(L-C)}{8L}$, where $L$ is the maximum possible concentration and $C(0) = 0.2$ mol/L is the initial concentration.

By solving the differential equation, show that $C = \frac{0.2 L}{(L-0.2)e^{-\frac{t}{8}}+0.2}$.

At $t=12$ minutes, the concentration of the product has reached $0.6$ mol/L.

Find the value of $L$, giving your answer correct to four significant figures.

Find the value of $t$ when the rate of change of the concentration is at its maximum.

The growth of a bacterial colony, $B$, in a petri dish can be modelled by the logistic differential equation

$\frac{\text{d}B}{\text{d}t} = k B \left(1 - \frac{B}{N}\right)$

where $t$ is the time measured in hours and $k, N$ are positive constants.

The constant $N$ represents the maximum number of bacteria the petri dish can sustain indefinitely due to limited nutrients.

In the context of this bacterial growth model, interpret the meaning of $\frac{\text{d}B}{\text{d}t}$.

Show that $\frac{\text{d}^2B}{\text{d}t^2} = k^2B\left(1-\frac{B}{N}\right)\left(1-\frac{2B}{N}\right)$.

Hence show that the bacterial colony will grow at its maximum rate when $B = \frac{N}{2}$. Justify your answer.

Hence determine the maximum value of $\frac{\text{d}B}{\text{d}t}$ in terms of $k$ and $N$.

Let $B_0$ be the initial number of bacteria.

By solving the logistic differential equation, show that its solution can be expressed in the form

$kt = \ln\left(\frac{B(N-B_0)}{B_0(N-B)}\right)$.

After 5 hours, the number of bacteria is $2B_0$. It is known that $N = 3B_0$.

Find the value of $k$ for this bacterial growth model.

(a) In a controlled biological experiment, the rate of change of the population $P$ of a certain microorganism with respect to time $t$ is modeled by the differential equation $t^2 \frac{dP}{dt} = P^2 - 2tP + 2t^2$, where $t > 0$ is in hours and $P$ is in thousands of organisms. It is known that at $t = 1$ hour, the population is $P = 4$ thousand.

Use Euler's method, with a step length of 0.1, to find an approximate value of $P$ when $t = 1.4$.

(b) Use the substitution $P = vt$ to show that $t\frac{dv}{dt} = v^2 - 3v + 2$.

(c.i) By solving the differential equation from part (b), and given that $P > 2t$, show that $P = \frac{6t - 2t^2}{3 - 2t}$.

(c.ii) Find the actual value of $P$ when $t = 1.4$.

(c.iii) Using the graph of $P = \frac{6t - 2t^2}{3 - 2t}$, suggest a reason why the approximation given by Euler's method in part (a) is not a good estimate to the actual value of $P$ at $t = 1.4$.

Consider the differential equation

$\frac{dy}{dx} = e^y \sin x \cos x, y \in \mathbb{R}$.

The solutions to this differential equation can be expressed in the form

$y = \ln\left(\frac{a}{\cos(2x) + b}\right)$,

where $a$ and $b$ are constants such that $a \in \mathbb{Z}^+$ and $b \in \mathbb{R}$.

(a) By solving the differential equation, find the value of $a$.

(b) Given that $y = 0$ when $x = \frac{\pi}{4}$, find the value of $b$.

Consider the differential equation $\frac{dy}{dx} = y^2 \sin x$.

Given that $y = 1$ when $x = 0$, show that the solution to the equation is $y = \sec x$.

Determine the value of the constant $A$ for which the following limit exists, and evaluate the limit:

$\lim_{x\to 0} \frac{\sec x - A}{x^2}$

A metal object is cooling in a room. Its temperature, $T$ (in degrees Celsius), at time $t$ (in minutes) can be modelled by the differential equation $\frac{dT}{dt} = -0.05(T-20)$. Initially, at $t=0$, the temperature of the object is $80^{\circ}\text{C}$.

(a) Use Euler's method with a step size of 0.1 to find an approximation for the temperature of the object when $t=0.4$ minutes. Give your answer correct to four significant figures.

(b) By solving the differential equation, show that $T = 20 + 60e^{-0.05t}$.

(c) Find the absolute value of the error in your approximation in part (a).

Find the general solution to the following differential equation. (a)

$\frac{dy}{dx} = \frac{y^2}{x^2-1}$

(b)

$\frac{dy}{dx} = y^2 \sin x \cos x$

(c) Find the particular solution to the differential equation $(x^2+4) \frac{dy}{dx} = xy$, given the initial condition $y(0)=1$.

(d)

$e^x \frac{dy}{dx} = \frac{1}{y}$

(e)

$\frac{dy}{dx} + 2y = xe^{-x}$

(f) $x \frac{dy}{dx} - 3y = x^5$ for $x>0$.

(g) $\frac{dy}{dx} + y \cot x = \cos x$ for $0 < x < \pi$.

This question explores families of curves and their intersections, including orthogonal trajectories and curves intersecting at a specific acute angle.

Consider a family of curves, $L$, with equation $xy = c$, where $c$ is a parameter. Each member of $L$ intersects every member of a family of curves, $C$, at right-angles.

Note: In parts (i), (ii) and (iii), you are not required to consider the case where $x = 0$ or $y = 0$.

Write down an expression for the gradient of $L$ in terms of $x$ and $y$.

Hence show that the gradient of $C$ is given by $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y}$.

By solving the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y}$, show that the family of curves, $C$, has equation $y^2 - x^2 = K$ where $K$ is a parameter.

Consider two families of curves: $F_1$ with equation $x^2 - y^2 = A$ and $F_2$ with equation $xy = B$. For this part, let $A = 3$ and $B = 2$.

On the same set of axes, sketch the curves $x^2 - y^2 = 3$ and $xy = 2$. On your sketch, clearly label each curve and any $x$-intercepts.

Find the coordinates of the intersection points of the curves $x^2 - y^2 = 3$ and $xy = 2$.

At the point $(2, 1)$, show that the curves $x^2 - y^2 = 3$ and $xy = 2$ intersect at right-angles.

Consider two families of curves, $F$ and $G$.

The gradient of $F$ is denoted by $f(x, y)$.

The gradient of $G$ is denoted by $g(x, y)$.

Each member of $F$ intersects every member of $G$ at an acute angle, $\alpha$.

It can be shown that

$g(x, y) = \frac{f(x, y) + \tan \alpha}{1 - f(x, y) \tan \alpha}$

In part (e), consider the specific case where $f(x, y) = -\frac{y}{x}$, for $x \neq 0$, $y \neq 0$ and $\alpha = \frac{\pi}{4}$.

Show that $g(x, y) = \frac{x-y}{x+y}$.

Hence, by solving the homogeneous differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x-y}{x+y}$, find a general equation that represents this family of curves, $G$. Give your answer in the form $h(x, y) = d$ where $d$ is a parameter.

By considering $\lim_{\alpha \to \frac{\pi}{2}} \tan \alpha$, show that, for all finite $f(x, y)$,

$\lim_{\alpha \to \frac{\pi}{2}} g(x, y) = -\frac{1}{f(x, y)}$.

(a) The concentration of an experimental drug in a patient's bloodstream, $C$, decreases at a rate proportional to its current concentration. Express this relationship as a differential equation.

(b) Solve the differential equation from part (a) to find an expression for $C$ in terms of time $t$, initial concentration $C_0$, and the constant of proportionality $k$.

(c) The half-life of the drug in the bloodstream is $4$ hours. Determine, to the nearest hour, how long it takes for the drug's concentration to fall to $15\%$ of its initial amount.

This question asks you to investigate the motion of a buoy bobbing up and down in the water.

A buoy bobs up and down in the water.

A fixed origin $\text{O}$ is the equilibrium position of the buoy (the water level).

The buoy's displacement, $y$ metres, from $\text{O}$ at time $t$ seconds is given by

$y = 6\sin\left(2t + \frac{\pi}{6}\right), \text{ for } 0 \le t \le \pi.$

Determine

the amplitude of the buoy's motion;

the buoy's initial displacement from $\text{O}$;

the value of $t$ when the buoy first passes through $\text{O}$.

Now consider the general case of a buoy bobbing up and down.

The buoy's acceleration is always directed towards a fixed origin $\text{O}$ at its equilibrium position.

The buoy's acceleration, $a$, at a displacement, $y$, from $\text{O}$ satisfies the differential equation

$a = -\omega^2 y, \text{ where } \omega > 0.$

The buoy's displacement, $y$, from $\text{O}$ at time $t$ is given by

$y = H\sin(\omega t + c), \text{ where } t \ge 0,\ H, \omega > 0 \text{ and } -\pi \le c \le \pi.$

By finding expressions for $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}t^2}$, verify that $y = H\sin(\omega t + c)$ satisfies the differential equation $a = -\omega^2 y$.

Use the chain rule to show that $a = v\frac{\mathrm{d}v}{\mathrm{d}y}$, where $v$ is velocity.

By solving the differential equation, $v\frac{\mathrm{d}v}{\mathrm{d}y} = -\omega^2 y$, show that $v^2 = \omega^2(H^2 - y^2)$.

Hence, or otherwise, find the buoy's maximum speed.

The continuous random variable $Y$ denotes the buoy's displacement, $y$, from $\text{O}$ at time $t$.

The probability density function $f$ of $Y$ is defined by

$f(y) = \begin{cases} \frac{1}{\pi\sqrt{H^2 - y^2}}, & -H < y < H \\ 0, & \text{otherwise.} \end{cases}$

Show that $\mathrm{P}\left(0 \le Y \le \frac{H\sqrt{3}}{2}\right) = \frac{1}{3}$.

For $-H < y < H$, the function $f(y)$ can be expressed in the form $\frac{m}{|v(y)|}$, where $m > 0$ and $v(y)$ is the buoy's velocity at a displacement, $y$, from $\text{O}$.

Find the value of $m$.

Determine $\mathrm{E}(Y)$, justifying your answer.

Interpret the result found in part (f)(i) in the context of the buoy's motion.