Complete solutions and integration choices | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Choose and carry out the integral

$\int \frac{f'(u)}{f(u)}\,du$ produces a logarithm.
$\int \frac{1}{a^2+u^2}\,du$ produces an inverse tangent.
A rational function may need polynomial division or partial fractions before integration.
Substitution and integration by parts remain available when their recognition cues appear.

Examiner note

The syllabus permits any integration technique from section 3.5 inside a separable differential equation. Separation does not replace those techniques.

Worked example

Separate, decompose, then integrate

Solve $\dfrac{dy}{dx}=xy(y+1)$ , given that $y=1$ when $x=0$ .

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

\frac{1}{y(y+1)}=\frac1y-\frac{1}{y+1}

\ln\left|\frac{y}{y+1}\right|=\frac{x^2}{2}+C

The condition gives $C=\ln(1/2)$ . On the positive branch:

\frac{y}{y+1}=\frac12e^{x^2/2}

Let $E=e^{x^2/2}$ and solve $2y=E(y+1)$ .

Answer

y=\dfrac{e^{x^2/2}}{2-e^{x^2/2}}

|x|<\sqrt{2\ln2}

Common mistake

Do not integrate

1/[y(y+1)]

as one logarithm. Its numerator is not the derivative of the whole denominator.

Check values removed by division

Do not lose solutions when separating

1. Before dividing

Solve g(y) = 0

Possible constant solutions.

↓

2. Then separate

Divide by g(y)

Solve the non-constant branch.

↓

3. Finish

Check in the original DE

Restore every valid solution.

Dividing by a y-factor can remove constant solutions, so test its zeros in the original differential equation.

Suppose $dy/dx=xy(y-2)$ . Dividing by $y(y-2)$ assumes $y\ne0$ and $y\ne2$ . Check those values first in the original equation: both $y=0$ and $y=2$ have zero derivative, so both are constant solutions.

For the non-constant branch:

\frac{1}{y(y-2)}\,dy=x\,dx

After integrating, multiply by 2 and rename the arbitrary constant:

-\ln|y|+\ln|y-2|=x^2+C

Answer

\ln\left|\dfrac{y-2}{y}\right|=x^2+C

\text{and }y=0,\ y=2

Key idea

This check matters when a general solution is required. A particular condition such as

y=1

already identifies a non-constant branch, but the division should still be understood.

Use an explicit or implicit answer as required

An explicit answer isolates y, such as $y=f(x)$ . An implicit answer leaves x and y in one relation. Use the form named by the question; rearranging an acceptable implicit relation can create unnecessary errors.

Your turnIndependent complete-solution check [7 marks]

Find the complete general solution of $\dfrac{dy}{dx}=xy(y+1)$ . Leave the non-constant branch as an implicit relation.

Show worked answer

Before dividing, $y=0$ and $y=-1$ both satisfy the original equation.

\frac{1}{y(y+1)}=\frac1y-\frac{1}{y+1}

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

Answer

\ln\left|\dfrac{y}{y+1}\right|=\dfrac{x^2}{2}+C

\text{and }y=0,\ y=-1

Differentiate the implicit relation to check the non-constant branch, and substitute each constant solution into the original equation.

(Pure Mathematics 2 & 3 Coursebook, Ch. 10.1)

Choose and carry out the integral

$\int \frac{f'(u)}{f(u)}\,du$ produces a logarithm.
$\int \frac{1}{a^2+u^2}\,du$ produces an inverse tangent.
A rational function may need polynomial division or partial fractions before integration.
Substitution and integration by parts remain available when their recognition cues appear.

Examiner note

The syllabus permits any integration technique from section 3.5 inside a separable differential equation. Separation does not replace those techniques.

Worked example

Separate, decompose, then integrate

Solve $\dfrac{dy}{dx}=xy(y+1)$ , given that $y=1$ when $x=0$ .

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

\frac{1}{y(y+1)}=\frac1y-\frac{1}{y+1}

\ln\left|\frac{y}{y+1}\right|=\frac{x^2}{2}+C

The condition gives $C=\ln(1/2)$ . On the positive branch:

\frac{y}{y+1}=\frac12e^{x^2/2}

Let $E=e^{x^2/2}$ and solve $2y=E(y+1)$ .

Answer

y=\dfrac{e^{x^2/2}}{2-e^{x^2/2}}

|x|<\sqrt{2\ln2}

Common mistake

Do not integrate

1/[y(y+1)]

as one logarithm. Its numerator is not the derivative of the whole denominator.

Check values removed by division

Do not lose solutions when separating

1. Before dividing

Solve g(y) = 0

Possible constant solutions.

↓

2. Then separate

Divide by g(y)

Solve the non-constant branch.

↓

3. Finish

Check in the original DE

Restore every valid solution.

Dividing by a y-factor can remove constant solutions, so test its zeros in the original differential equation.

For the non-constant branch:

\frac{1}{y(y-2)}\,dy=x\,dx

After integrating, multiply by 2 and rename the arbitrary constant:

-\ln|y|+\ln|y-2|=x^2+C

Answer

\ln\left|\dfrac{y-2}{y}\right|=x^2+C

\text{and }y=0,\ y=2

Key idea

This check matters when a general solution is required. A particular condition such as

y=1

already identifies a non-constant branch, but the division should still be understood.

Use an explicit or implicit answer as required

Your turnIndependent complete-solution check [7 marks]

Find the complete general solution of $\dfrac{dy}{dx}=xy(y+1)$ . Leave the non-constant branch as an implicit relation.

Show worked answer

Before dividing, $y=0$ and $y=-1$ both satisfy the original equation.

\frac{1}{y(y+1)}=\frac1y-\frac{1}{y+1}

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

Answer

\ln\left|\dfrac{y}{y+1}\right|=\dfrac{x^2}{2}+C

\text{and }y=0,\ y=-1

Differentiate the implicit relation to check the non-constant branch, and substitute each constant solution into the original equation.

(Pure Mathematics 2 & 3 Coursebook, Ch. 10.1)