First Order Equations (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Solving using exact equations

Question: Solve the differential equation $x^2\frac{dy}{dx} + 2xy = 4x^3$, giving your answer in the form $y = f(x)$.

Solution:

The left-hand side can be recognised as the derivative of a product. Specifically, it matches the pattern for $\frac{d}{dx}[x^2y]$.

Step 1: Recognise the left side as a product derivative: $\frac{d}{dx}[x^2y] = 4x^3$

Step 2: Integrate both sides: $x^2y = x^4 + c$

Step 3: Rearrange to express $y$ in terms of $x$: $y = \frac{x^4 + c}{x^2} = x^2 + \frac{c}{x^2}$

This example shows how recognizing the product rule pattern allows immediate integration without needing an integrating factor.

Worked Example 2: Using an integrating factor

Question: Solve the differential equation $\frac{dy}{dx} - y\tan x = 3\sin^2 x$, for $0 \leq x < \frac{\pi}{2}$.

Solution:

This equation is already in standard form with $P(x) = -\tan x$ and $f(x) = 3\sin^2 x$.

Step 1: Find the integrating factor: $e^{\int P(x)dx} = e^{\int -\tan x dx} = e^{-\ln(\cos x)} = \cos x$

Step 2: Multiply through by the integrating factor: $\cos x \frac{dy}{dx} - y\cos x \tan x = 3\cos x \sin^2 x$

Since $\cos x \tan x = \sin x$: $\cos x \frac{dy}{dx} - y\sin x = 3\cos x \sin^2 x$

Step 3: Recognise this is now an exact equation: $\frac{d}{dx}[y\cos x] = 3\cos x \sin^2 x$

Step 4: Integrate both sides: $y\cos x = \sin^3 x + c$

You can leave the solution in implicit form: $y\cos x = \sin^3 x + c$.

Worked Example 3: Finding particular solutions

Question: a) Find the general solution to the equation $x\frac{dy}{dx} - y = 2x^3$

b) Given that $y = 4$ when $x = 1$, find the particular solution.

Solution:

Part a: First, rearrange into standard form: $\frac{dy}{dx} - \frac{1}{x}y = 2x^2$

So $P(x) = -\frac{1}{x}$ and the integrating factor is: $e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \frac{1}{x}$

Multiply throughout by $\frac{1}{x}$: $\frac{1}{x}\frac{dy}{dx} - \frac{1}{x^2}y = 2x$

Recognise the exact equation: $\frac{d}{dx}\left[\frac{1}{x}y\right] = 2x$

Integrate both sides: $\frac{y}{x} = x^2 + c$

Therefore, the general solution is: $y = x^3 + cx$

Part b: Substitute $y = 4$ and $x = 1$: $4 = 1^3 + c(1)$ $c = 3$

Therefore, the particular solution is: $y = x^3 + 3x$

This example demonstrates how initial conditions allow you to find a specific value for the constant of integration.

Worked Example 4: Using substitutions

Question: a) Use the substitution $y = u^2$ to transform the differential equation $\frac{dy}{dx} + x^2y = x\sqrt{y}$ into the differential equation $\frac{du}{dx} + \frac{1}{2}x^2u = \frac{1}{2}x^2$

b) Hence use an integrating factor to find the general solution to the differential equation $\frac{dy}{dx} + x^2y = x\sqrt{y}$ in the form $y = f(x)$.

Solution:

Part a: Find an expression for $\frac{dy}{dx}$ using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 2u\frac{du}{dx}$

Now substitute into the original equation. Since $y = u^2$ and $\sqrt{y} = u$: $2u\frac{du}{dx} + x^2u^2 = xu$

Divide through by $2u$: $\frac{du}{dx} + \frac{1}{2}x^2u = \frac{1}{2}x^2$

Part b: This is now in standard form with $P(x) = \frac{1}{2}x^2$.

Find the integrating factor: $e^{\int \frac{1}{2}x^2 dx} = e^{\frac{x^3}{6}}$

Multiply through by the integrating factor: $e^{\frac{x^3}{6}}\frac{du}{dx} + \frac{1}{2}x^2e^{\frac{x^3}{6}}u = \frac{1}{2}x^2e^{\frac{x^3}{6}}$

The left side is the derivative of $e^{\frac{x^3}{6}}u$: $e^{\frac{x^3}{6}}u = \int \frac{1}{2}x^2e^{\frac{x^3}{6}}dx = e^{\frac{x^3}{6}} + c$

Therefore: $e^{\frac{x^3}{6}}\sqrt{y} = e^{\frac{x^3}{6}} + c$

Substitute back for $y$ and make $y$ the subject: $\sqrt{y} = 1 + ce^{-\frac{x^3}{6}}$

$y = \left(1 + ce^{-\frac{x^3}{6}}\right)^2$

Notice how the substitution transformed a complex equation into one solvable by standard methods.

Worked Example 5: Application to population growth

Question: A population of bacteria has an initial size of 100. After $t$ hours, the size of the population is $P$. The connection between $P$ and $t$ can be modelled by the equation $\frac{dP}{dt} = 2(50t - P)$.

a) Solve this equation to show that $P = 25(2t + 5e^{-2t} - 1)$

b) Find the size of the population after 24 hours.

c) Prove that the number of bacteria never falls below 40.

Solution:

Part a: Rearrange into standard form: $\frac{dP}{dt} + 2P = 100t$

The integrating factor is: $e^{\int 2dt} = e^{2t}$

Multiply through by $e^{2t}$: $e^{2t}\frac{dP}{dt} + 2e^{2t}P = 100te^{2t}$

The left side becomes: $\frac{d}{dt}[e^{2t}P] = 100te^{2t}$

Integrate using integration by parts: $e^{2t}P = 50te^{2t} - \int 50e^{2t}dt = 50te^{2t} - 25e^{2t} + c$

Use initial conditions: when $t = 0$, $P = 100$:

$100 = 0 - 25 + c$

$c = 125$

Therefore: $e^{2t}P = 50te^{2t} - 25e^{2t} + 125$

$P = 25(2t + 5e^{-2t} - 1)$

Part b: Substitute $t = 24$: $P = 25[2(24) + 5e^{-2(24)} - 1] = 25[48 + 5e^{-48} - 1] \approx 1175$

The population after 24 hours is approximately 1175 bacteria.

Part c: To find the minimum value, set $\frac{dP}{dt} = 0$:

$2(50t - P) = 0$

$50t = P$

From the solution: $P = 25(2t + 5e^{-2t} - 1)$

Setting equal:

$50t = 25(2t + 5e^{-2t} - 1)$

$2t = 2t + 5e^{-2t} - 1$

$5e^{-2t} = 1$

$t = \frac{1}{2}\ln 5$

At this time: $P = 25(\ln 5 + 5 - 1) = 25(\ln 5 + 4) \approx 40.2$

By inspection, this is a minimum (as $P$ increases before and after this point), so the value of $P$ never falls below approximately 40.

Worked Example 6: Raindrop velocity

Question: A raindrop falls vertically through a cloud. Initially the raindrop is at rest. At time $t$ seconds, the velocity $v$ m s⁻¹ of the raindrop satisfies the equation $(2+t)\frac{dv}{dt} + v = 10(2+t)$.

a) Solve this equation to find an expression for $v$ in terms of $t$

b) Find the time at which the velocity of the raindrop is 21 m s⁻¹

c) Criticise the model.

Solution:

Part a: Rearrange into standard form: $\frac{dv}{dt} + \frac{1}{2+t}v = 10$

The integrating factor is: $e^{\int \frac{1}{2+t}dt} = e^{\ln(2+t)} = (2+t)$

Multiply throughout by $(2+t)$: $(2+t)\frac{dv}{dt} + v = 10(2+t)$

Note that the equation was already in the form $[u'v + uv']$, where the integrating factor is $(2+t)$.

The left side is: $\frac{d}{dt}[(2+t)v] = 10(2+t)$

Integrate: $(2+t)v = 5(2+t)^2 + c$

Use initial conditions: when $t = 0$, $v = 0$:

$0 = 5(4) + c$

$c = -20$

Therefore: $v = \frac{5(2+t)^2 - 20}{2+t} = 5(2+t) - \frac{20}{2+t}$

Part b: Set $v = 21$: $21 = 5(2+t) - \frac{20}{2+t}$

$21 = 10 + 5t - \frac{20}{2+t}$

Multiply through by $(2+t)$: $21(2+t) = (10 + 5t)(2+t) - 20$

$42 + 21t = 20 + 10t + 10t + 5t^2 - 20$

$5t^2 - t - 42 = 0$

$(5t + 14)(t - 3) = 0$

So $t = 3$ seconds (taking the positive value).

Part c: Under this model, as $t$ increases, $v$ becomes unlimited. However, in practice, the raindrop would approach a terminal velocity due to air resistance. The model fails to account for this physical limitation.

Worked Example 7: Rocket velocity

Question: A fuel-filled rocket of mass 25 kg burns fuel at a rate of 2 kg s⁻¹. The rocket is initially at rest. After $t$ seconds, the velocity $v$ m s⁻¹ of the rocket satisfies the equation $(25-2t)\frac{dv}{dt} + v = 100$ for $t \leq 12$.

a) Show that $v = 100 - 20\sqrt{25-2t}$

b) Find the speed of the rocket when $t = 12$

c) Sketch a graph of $v$ against $t$

d) What happens for $t > 12.5$?

Solution:

Part a: Rearrange into standard form: $\frac{dv}{dt} + \frac{1}{25-2t}v = \frac{100}{25-2t}$

The integrating factor is: $e^{\int \frac{1}{25-2t}dt} = e^{-\frac{1}{2}\ln(25-2t)} = (25-2t)^{-\frac{1}{2}}$

Multiply throughout by $(25-2t)^{-\frac{1}{2}}$: $(25-2t)^{-\frac{1}{2}}\frac{dv}{dt} + (25-2t)^{-\frac{3}{2}}v = 100(25-2t)^{-\frac{3}{2}}$

The left side is: $\frac{d}{dt}\left[(25-2t)^{-\frac{1}{2}}v\right] = 100(25-2t)^{-\frac{3}{2}}$

Integrate: $(25-2t)^{-\frac{1}{2}}v = 100(25-2t)^{-\frac{1}{2}} + c$

Use initial conditions: when $t = 0$, $v = 0$: $0 = \frac{100}{\sqrt{25}} + c$ $c = -20$

Therefore: $(25-2t)^{-\frac{1}{2}}v = 100(25-2t)^{-\frac{1}{2}} - 20$

$v = 100 - 20\sqrt{25-2t}$

Part b: When $t = 12$: $v = 100 - 20\sqrt{25-2(12)} = 100 - 20\sqrt{1} = 80 \text{ m s}^{-1}$

The rocket has a speed of 80 m s⁻¹ when $t = 12$ seconds.

Part c: The graph shows velocity increasing from 0 at $t = 0$ in a curved manner, reaching 80 m s⁻¹ at $t = 12$.

Part d: For $t > 12.5$, the square root becomes negative, so the solution does not make sense physically. The rocket has burned all of its fuel by $t = 12.5$, so the equation no longer applies. A different model would be needed to describe the motion after this point.