Particular Solutions Revision Notes for AQA A-Level Mathematics

8.3.2 Particular Solutions

Differential Equations

A differential equation is an equation involving a differential, such as $\dfrac{dy}{dx}$ . To solve a differential equation is to eliminate the differential by integration.

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Example

Solve $\frac{dy}{dx} = x^2 + 3x - 2$

Integrating both sides,

\int \frac{dy}{dx} dx = \int (x^2 + 3x - 2) dx

\Rightarrow y = \frac{x^3}{3} + \frac{3x^2}{2} - 2x + c

This is called a general solution because it contains a " $+c$ ". If we were given values to sub in to find the value of $c$ , this would be called a particular solution.

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Example: Given that $y = f(x)$ passes through $(4, 7)$ , and $\frac{dy}{dx} = \sqrt{x} + 3$ , find a particular solution for $y$ in terms of $x$ .

\frac{dy}{dx} = {x}^{\frac {1}{2}} + 3 \quad \Rightarrow \quad \int \frac{dy}{dx} dx = \int ({x}^{\frac {1}{2}} + 3) dx

\Rightarrow \frac{2}{3}x^{\frac{3}{2}} + 3x + c

Now subbing in the given point $x = 4, y = 7$ :

7 = \frac{2}{3}x^{\frac{3}{2}} + 3x + c

7 = \frac{2}{3}(4)^{\frac{3}{2}} + 3(4) + c

c = -\frac{31}{3}

\Rightarrow y = \frac{2}{3}x^{\frac{3}{2}} + 3x - \frac{31}{3}

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The gradient of a curve is given by $\frac{dy}{dx} = 3x^2 + a$ , where $a$ is a constant. The curve passes through the points $(-1, 2)$ and $(2, 17)$ . Find the equation of the curve.

$y = \int 3x^2 + a \, dx \Rightarrow y = x^3 + axe + c$
Using the point $(-1, 2)$ :

\text{Let } x = -1, y = 2 \Rightarrow 2 = (-1)^3 + a(-1) + c

\Rightarrow 2 = -1 - a + c \\\Rightarrow -a + c = 3

Using the point $(2, 17)$ :

\text{Let } x = 2, y = 17 \Rightarrow 17 = (2)^3 + a(2) + c

\Rightarrow 17 = 8 + 2a + c \Rightarrow 2a + c = 9

Solving the equations:

a + c = 3 \quad \text{and} \quad 2a + c = 9

\Rightarrow a = 2, c = 5

Therefore, the equation of the curve is:

y = x^3 + 2x + 5

(i) Find $\int (6x^{\frac{1}{2}} - 1) \, dx$ .

\int (6x^{\frac{1}{2}} - 1) \, dx = \frac {2}{3}\times 6x^{\frac{3}{2}}-x+c

\Rightarrow 4x^{\frac{3}{2}} - x + c

(ii) Hence find the equation of the curve for which $\frac {dy}{dx} = 6x^{\frac {1}{2}} - 1$ and which passes through the point $(4, 17)$ .

$y = 4x^{\frac{3}{2}} - x + c$
Using the point $(4, 17)$ :

\text{Let } x = 4, y = 17 \Rightarrow 17 = 4(4)^{\frac{3}{2}} - 4 + c

\Rightarrow 17 = 28 + c \Rightarrow c = -11

Therefore, the equation of the curve is:

y = 4x^{\frac{3}{2}} - x - 11

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The gradient of a curve is given by $\frac {dy}{dx} = 6x - 4$ . The curve passes through the distinct points $(2, 5)$ and ( $p$ , $5$ ).

(i) Find the equation of the curve.

\frac {dy}{dx} = 6x - 4 \Rightarrow y = \int (6x - 4) \, dx = \frac {6x^2}{2} - 4x + c

\therefore \quad y = 3x^2 - 4x + c

Using the point $(2, 5)$ :

\text{Let } x = 2, y = 5 \Rightarrow 5 = 3(2)^2 - 4(2) + c

\Rightarrow 5 = 4 + c \Rightarrow c = 1

Therefore, the equation of the curve is:

y = 3x^2 - 4x + 1

(ii) Find the value of $p$ . Using the point ( $p$ , $5$ ):

\text{Let } x = p, y = 5 \Rightarrow 5 = 3p^2 - 4p + 1

\Rightarrow 3p^2 - 4p - 4 = 0

p = \xcancel2 \quad \text{or} \quad p = -\frac{2}{3}

However, since $p = 2$ is already given in the question (i.e., distinct points $(2, 5)$ and ( $p$ , $5$ ), the only solution for p that maintains distinct points is:

p = -\frac{2}{3}

(The solution $p = 2$ is discarded because it would not provide distinct points.)

Differential Equations

A differential equation is an equation containing a differential.

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e.g., $\frac{dy}{dx} = x^2 - 2$ .

Solving a differential equation means eliminating the differential, i.e., in this case, write only in terms of $y$ and $x$ .
A general solution to a differential equation is one that contains $a$ constant $c$ .

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e.g., Find the general solution of $\frac{dy}{dx} = x^2 - 2$ :

\int \frac{dy}{dx} dx = \int (x^2 - 2) dx

\Rightarrow y = \frac{1}{3}x^3 - 2x + c

(This $c$ makes the solution general.)

A particular solution to a differential equation is where we find the value of $c$ .

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e.g., Given that the above example curve passes through $(0, 3)$ , find the particular solution for these conditions: Let $x = 0, y = 3$ :

3 = \frac{1}{3}(0)^3 - 2(0) + c \Rightarrow c = 3

\Rightarrow y = \frac{1}{3}x^3 - 2x + 3

(This $+3$ makes the solution particular.)

The Method of Separating The Variables

In the previous example, the differential equation was of the form $\frac{dy}{dx} = f(x)$ , which made it easy to integrate. Often we find that the RHS of differential equations are a mixture of $x's$ and $y's$ making them more difficult to integrate.

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Example: Find the general solution to $\frac{dy}{dx} = xy$ .

Split up the $dy$ and $dx$ by multiplying by $dx$ . This tells us which side we need the $x's$ and $y's$ on.

dy = xy \, dx \quad (x dx)

Get the $x's$ on the same side as the $dx$ and the $y's$ on the same side as the $dy$ .

\frac{1}{y} \, dy = x \, dx \quad (\div y)

Integrate both sides:

\int \frac{1}{y} \, dy = \int x \, dx

\ln|y| = \frac{1}{2}x^2 + c

(Constant only needed on one side)

Note: At this point, we have fulfilled the requirements of the question, i.e., eliminated the differential.

:::

Suppose instead the question asked:

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Find the general solution for $y$ to $\frac{dy}{dx} = xy$ . The words "for $y$ " mean write in the form $y = f(x)$ .

So:

\ln|y| = \frac{1}{2}x^2 + c

\div e^\square

e^{\ln|y|} = e^{\frac{1}{2}x^2 + c}

y = e^{\frac{1}{2}x^2} \times e^c

(Just a number so can be represented by any letter)

\therefore \quad y = Ae^{\frac{1}{2}x^2}

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Example: Find the general solution of $(x + 1) \frac{dy}{dx} = y$ .

(x + 1) \, dy = y \, dx \quad (\times dx)

\frac{1}{y} \, dy = \frac{1}{x+1} \, dx \quad (\div (x+1) \text{ and then } \div y)

\int \frac{1}{y} \, dy = \int \frac{1}{x+1} \, dx \quad (\int \text{ both sides})

\Rightarrow \ln|y| = \ln|x+1| + c

(Didn't ask for $y = f(x)$ $\therefore$ done)

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Example: Find the particular solution to $\frac{dy}{dx} = x^2 \tan y$ , given that $x = 0$ when $y = \frac{\pi}{6}$ .

dy = x^2 \tan y \, dx

\frac{1}{\tan y} \, dy = x^2 \, dx

\cot y \, dy = x^2 \, dx

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

Integration

\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c

Notice that LHS is in the above form: $\int \frac{\cos y}{\sin y} \, dy = \ln|\sin y| + c$

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

\int LHS dy = \ln|\sin y|+c

\int RHS dy = \frac{1}{3}x^3 + c

\therefore \quad \ln|\sin y| = \frac{1}{3}x^3 + c

We were asked to find the particular solution, so we must find the value of $c$ .

Let $x = 0, y = \frac{\pi}{6}$

\Rightarrow \ln \left| \sin \frac{\pi}{6} \right| = \frac{1}{3}(0)^3 + c

\Rightarrow \ln \left( \frac{1}{2} \right) = c

\Rightarrow \ln|\sin y| = \frac{1}{3} x^3 + \ln \left( \frac{1}{2} \right)

(or -\ln 2)

Differential Equations in Context

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Q3. (Jan 2007, Q8) The height, $h$ metres, of a shrub $t$ years after planting is given by the differential equation:

\frac{dh}{dt} = \frac{6 - h}{20}

A shrub is planted when its height is 1 m.

$h = 1$ when $t = 0$ .

i) Show by integration that $t = 20 \ln \left(\frac{5}{6 - h}\right)$ .

ii) How long after planting will the shrub reach a height of 2 m?

iii) Find the height of the shrub 10 years after planting.

iv) State the maximum possible height of the shrub.

SOLUTION:

(i)

\frac{dh}{dt} = \frac{6-h}{20}

\Rightarrow dh = \frac{6-h}{20} dt

Note: Only rearrange what is absolutely necessary

\Rightarrow \frac{1}{6-h} dh = \frac{1}{20} dt \quad \left(\div (6-h)\right)

\Rightarrow \int \frac{1}{6-h} dh = \int \frac{1}{20} dt \quad \quad \int

\Rightarrow \ln|6-h| = \frac{1}{20}t + c

Remember to $\div$ by differential of $6 - h$

Info given in question: Let $h = 1, t = 0$ .

\Rightarrow -\ln(6-1) = \frac{1}{20}(0) + c \Rightarrow c = -\ln(5)

\Rightarrow -\ln(6-h) = \frac{t}{20} - \ln(5) \quad \left( \text{Adding } \ln(5) \right)

\Rightarrow \ln(5) - \ln(6-h) = \frac{t}{20}

\Rightarrow \ln\left(\frac{5}{6-h}\right) = \frac{t}{20} \Rightarrow t = 20\ln\left(\frac{5}{6-h}\right)

ii) Let $h = 2$ :

t = 20\ln\left(\frac{5}{6-2}\right) = 20\ln\left(\frac{5}{4}\right) \approx :highlight[4.463 \text{ years}]

iii) Let $t = 10$ :

10 = 20 \ln\left(\frac{5}{6-h}\right)

\Rightarrow \frac{1}{2} = \ln\left(\frac{5}{6-h}\right) \Rightarrow e^{\frac{1}{2}} = \frac{5}{6-h}

\Rightarrow 6-h = \frac{5}{e^{\frac{1}{2}}} \Rightarrow h = 6 - \frac{5}{e^{\frac{1}{2}}} \approx :highlight[2.967 \text{ m}]

iv) $h = :highlight[6] \quad \text{(since } h = 6 \text{ "breaks the equation")}$

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Q4. (Jan 2008, Q8)

Water flows out of a tank through a hole in the bottom and, at time $t$ minutes, the depth of water in the tank is $x$ metres. At any instant, the rate at which the depth of water in the tank is decreasing is proportional to the square root of the depth of water in the tank.

i) Write down a differential equation which models this situation.

\frac{dx}{dt} = -k\sqrt{x}

Decreasing rate $-$

Key Point $a \propto b \Rightarrow a = kb$

ii) When $t = 0, x = 2$ ; when $t = 5, x = 1$ . Find $t$ when $x = 0.5$ , giving your answer correct to 1 decimal place.

dx= -k\sqrt{x}

\Rightarrow \frac{1}{\sqrt{x}}dx = -k dt

\Rightarrow \int x^{-\frac{1}{2}} dx = -k \int dt \Rightarrow \int x^{-\frac{1}{2}} dx = \int -k dt

\Rightarrow 2x^{\frac{1}{2}} = -kt + c

Let $t = 0, x = 2 \Rightarrow 2\sqrt{2} = c$
Let $t = 5, x = 1 \Rightarrow 2\sqrt 1 = -5k + 2\sqrt{2}$ Find $k$ and $c$ using given points:

\Rightarrow \frac{2 - 2\sqrt{2}}{-5} = k \Rightarrow k = \frac{2\sqrt{2} - 2}{5}

\Rightarrow 2x^{\frac{1}{2}} = -\left(\frac{2\sqrt{2} - 2}{5}\right) t + 2\sqrt{2}

Let $x = 0.5$ :

\Rightarrow 2(0.5)^{\frac{1}{2}} = -\left(\frac{2\sqrt{2} - 2}{5}\right) t + 2\sqrt{2}

\Rightarrow \frac {2(0.5)^{\frac {1}{2}} - 2\sqrt{2}}{-\left (\frac{2\sqrt{2} - 2}{5}\right )} = t = :highlight[8.5 \text{ minutes}]

Particular Solutions (AQA A-Level Mathematics): Revision Notes

8.3.2 Particular Solutions

Differential Equations

Example

Differential Equations

The Method of Separating The Variables

Example: Find the general solution to $\frac{dy}{dx} = xy$ .

Example: Find the general solution of $(x + 1) \frac{dy}{dx} = y$ .

Example: Find the particular solution to $\frac{dy}{dx} = x^2 \tan y$ , given that $x = 0$ when $y = \frac{\pi}{6}$ .

Differential Equations in Context

Q4. (Jan 2008, Q8)

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Further Integration

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Particular Solutions (AQA A-Level Mathematics): Revision Notes

8.3.2 Particular Solutions

Differential Equations

Example

Differential Equations

The Method of Separating The Variables

Example: Find the general solution to dydx=xy\frac{dy}{dx} = xydxdy​=xy.

Example: Find the general solution of (x+1)dydx=y(x + 1) \frac{dy}{dx} = y(x+1)dxdy​=y.

Example: Find the particular solution to dydx=x2tan⁡y\frac{dy}{dx} = x^2 \tan ydxdy​=x2tany, given that x=0x = 0x=0 when y=π6y = \frac{\pi}{6}y=6π​.

Differential Equations in Context

Q4. (Jan 2008, Q8)

Explore AQA A-Level Mathematics Model Answers by Topics

Integration

Further Integration

Differential Equations

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Integration

Further Integration

Differential Equations

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Integration

Further Integration

Differential Equations

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Example: Find the general solution to $\frac{dy}{dx} = xy$ .

Example: Find the general solution of $(x + 1) \frac{dy}{dx} = y$ .

Example: Find the particular solution to $\frac{dy}{dx} = x^2 \tan y$ , given that $x = 0$ when $y = \frac{\pi}{6}$ .