Paper 2 · Further Pure 2 ✦

Differential Equations

An equation whose unknown is a whole function. Multiply by the right factor, guess the right shape, and the answer assembles itself out of two pieces — the complementary function and a particular integral.

Syllabus 2.6Nothing given by MF19General solution = CF + PI

What the examiner expects you to do

Find an integrating factor for a first order linear equation and use it to find the general solution.
Know what complementary function and particular integral mean, and that the general solution is their sum.
Find the CF for a first or second order linear equation with constant coefficients — distinct real roots, a repeated real root, or conjugate complex roots.
Recall the form of, and find, a PI when a polynomial, $a e^{b x}$ , or $a \cos p x + b \sin p x$ is suitable — and in other simple cases find the coefficients given a suitable form.
Use a given substitution to reduce an equation to linear-with-constant-coefficients form, or to a first order separable one.
Use initial conditions to find a particular solution, and interpret a solution in the context of a model.

§1 First order linear: the integrating factor

◆ The method

Put the equation into the standard form — coefficient of $\frac{d y}{d x}$ equal to $1$ :

\frac{d y}{d x} + P (x) y = Q (x) .

Multiply through by the integrating factor $I = e^{\int P d x}$ . The left-hand side then collapses into a single derivative:

\frac{d}{d x} (I y) = I Q ⟹ I y = \int I Q d x .

No constant of integration is needed when finding $I$ itself; one arbitrary constant appears at the final integration, which is exactly the one the general solution should have.

Why

e^{\int P d x}

is the magic multiplier

We want a function $I$ making $I \frac{d y}{d x} + I P y$ the derivative of $I y$ . By the product rule,

\frac{d}{d x} (I y) = I \frac{d y}{d x} + \frac{d I}{d x} y,

so we need $\frac{d I}{d x} = I P$ . That is separable: $\int \frac{d I}{I} = \int P d x$ , giving $\ln I = \int P d x$ and $I = e^{\int P d x}$ . $■$

✎ Worked example — a polynomial right-hand side

Solve $\frac{d y}{d x} - 2 y = x^{2}$ .

Already in standard form with $P = - 2$ , so $I = e^{- 2 x}$ and

\frac{d}{d x} (y e^{- 2 x}) = x^{2} e^{- 2 x} .

Integrate the right-hand side by parts twice:

\int x^{2} e^{- 2 x} d x = - \frac{1}{2} x^{2} e^{- 2 x} - \frac{1}{2} x e^{- 2 x} - \frac{1}{4} e^{- 2 x} + A = - \frac{1}{4} e^{- 2 x} (2 x^{2} + 2 x + 1) + A .

Hence $y e^{- 2 x} = - \frac{1}{4} e^{- 2 x} (2 x^{2} + 2 x + 1) + A$ , and multiplying by $e^{2 x}$ :

y = - \frac{1}{4} (2 x^{2} + 2 x + 1) + A e^{2 x} .

Check: $y^{'} = - x - \frac{1}{2} + 2 A e^{2 x}$ , so $y^{'} - 2 y = - x - \frac{1}{2} + 2 A e^{2 x} + \frac{1}{2} (2 x^{2} + 2 x + 1) - 2 A e^{2 x} = x^{2}$ ✓

✎ Worked example — divide through first

Solve $x \frac{d y}{d x} - y = x^{4}$ .

The coefficient of $\frac{d y}{d x}$ is $x$ , not $1$ , so divide first: $\frac{d y}{d x} - \frac{y}{x} = x^{3}$ . Then $P = - \frac{1}{x}$ , $\int P d x = - \ln x$ , and

I = e^{- \ln x} = \frac{1}{x} .

So $\frac{d}{d x} (\frac{y}{x}) = x^{2}$ , giving $\frac{y}{x} = \frac{x^{3}}{3} + A$ and

y = \frac{x^{4}}{3} + A x .

Check: $x y^{'} - y = x (\frac{4}{3} x^{3} + A) - \frac{1}{3} x^{4} - A x = \frac{4}{3} x^{4} - \frac{1}{3} x^{4} = x^{4}$ ✓ Forgetting to divide through is the most common error in this whole section — the integrating factor would have been wrong from the first line.

✎ Worked example — a hyperbolic integrating factor

Solve $\frac{d y}{d x} + y \coth x = \cosh x$ .

$P = \coth x = \frac{\cosh x}{\sinh x}$ , whose integral is $\ln | \sinh x |$ , so $I = \sinh x$ . Then

\frac{d}{d x} (y \sinh x) = \sinh x \cosh x ⟹ y \sinh x = \frac{1}{2} \sinh^{2} x + A,

because $\sinh x \cosh x$ is exactly $\frac{1}{2} \frac{d}{d x} (\sinh^{2} x)$ . Dividing,

y = \frac{1}{2} \sinh x + A cosech x .

Spotting that $\int P d x$ is a logarithm — so that $I$ comes out as a clean function rather than an exponential — is what makes these tractable.

§2 Complementary function and particular integral

◇ The two words, and why the sum works

For a linear equation $L (y) = f (x)$ :

the complementary function (CF) is the general solution of the homogeneous equation $L (y) = 0$ — it carries the arbitrary constants;
a particular integral (PI) is any one solution of the full equation $L (y) = f (x)$ — no arbitrary constants.

general solution = CF + PI

Why: if $y_{p}$ is a particular integral and $y$ is any solution, then $L (y - y_{p}) = L (y) - L (y_{p}) = f - f = 0$ , so $y - y_{p}$ is a complementary function. Every solution is therefore a CF plus that one PI — and none is missed.

⚠ Count your arbitrary constants

A first order equation has one arbitrary constant; a second order equation has two. If your final answer has the wrong number, something has been dropped — most often the CF itself.

§3 Finding the complementary function

◆ The auxiliary equation

For $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 0$ , try $y = e^{m x}$ . Substituting gives $(a m^{2} + b m + c) e^{m x} = 0$ , and since $e^{m x} \neq 0$ we need the auxiliary equation

a m^{2} + b m + c = 0 .

Roots	Complementary function	Behaviour
real and distinct, $m_{1} \neq m_{2}$	$y = A e^{m_{1} x} + B e^{m_{2} x}$	pure growth / decay
real and repeated, $m$	$y = (A + B x) e^{m x}$	the extra $x$ is essential — two constants are needed
complex, $p \pm q i$	$y = e^{p x} (A \cos q x + B \sin q x)$	oscillation inside an envelope $e^{p x}$

For a first order equation $a \frac{d y}{d x} + b y = 0$ the auxiliary equation is just $a m + b = 0$ , giving $y = A e^{- b x / a}$ .

✎ Worked examples — one of each

Distinct real: $y^{''} - 3 y^{'} + 2 y = 0$ has $m^{2} - 3 m + 2 = (m - 1) (m - 2) = 0$ , so $y = A e^{x} + B e^{2 x}$ .
Repeated: $y^{''} - 4 y^{'} + 4 y = 0$ has $(m - 2)^{2} = 0$ , so $y = (A + B x) e^{2 x}$ . Check the $B x$ term: if $y = x e^{2 x}$ then $y^{'} = (1 + 2 x) e^{2 x}$ and $y^{''} = (4 + 4 x) e^{2 x}$ , so $y^{''} - 4 y^{'} + 4 y = (4 + 4 x - 4 - 8 x + 4 x) e^{2 x} = 0$ ✓
Complex: $y^{''} + 2 y^{'} + 5 y = 0$ has $m = \frac{- 2 \pm \sqrt{4 - 20}}{2} = - 1 \pm 2 i$ , so $y = e^{- x} (A \cos 2 x + B \sin 2 x)$ — an oscillation of period $π$ damped by $e^{- x}$ .

★ A* tip — read the physics off the roots

The real part of the roots controls growth or decay, and the imaginary part controls oscillation. Negative real part means the solution dies away, whatever the constants — which is exactly the answer to "describe the long-term behaviour" questions.

§4 Finding a particular integral

◆ Trial forms — match the shape of the right-hand side

Right-hand side $f (x)$	Try
a polynomial of degree $n$	a general polynomial of degree $n$
$a e^{b x}$	$C e^{b x}$
$a \cos p x + b \sin p x$ (or just one of them)	$C \cos p x + D \sin p x$ — always both

Substitute the trial form into the equation and compare coefficients to pin down the constants.

⚠ Resonance: when the trial form is already in the CF

If your trial PI is a solution of the homogeneous equation, substituting it gives $0 = f (x)$ — nonsense. The remedy is to multiply the trial form by $x$ (and by $x^{2}$ if it is still in the CF, which happens with a repeated root).

✎ Worked example — the ordinary case

Solve $y^{''} - 3 y^{'} + 2 y = e^{3 x}$ .

CF: $m^{2} - 3 m + 2 = 0$ gives $m = 1,2$ , so CF $= A e^{x} + B e^{2 x}$ .

PI: $e^{3 x}$ is not in the CF, so try $y = C e^{3 x}$ . Then $y^{'} = 3 C e^{3 x}$ and $y^{''} = 9 C e^{3 x}$ , so

9 C - 9 C + 2 C = 1 ⟹ C = \frac{1}{2} .

General solution: $y = A e^{x} + B e^{2 x} + \frac{1}{2} e^{3 x}$ .

✎ Worked example — the resonant case

Solve $y^{''} - 3 y^{'} + 2 y = e^{2 x}$ . Same CF, but now $e^{2 x}$ is in it, so try $y = C x e^{2 x}$ :

y^{'} = C (1 + 2 x) e^{2 x}, y^{''} = C (4 + 4 x) e^{2 x},

y^{''} - 3 y^{'} + 2 y = C [(4 + 4 x) - 3 (1 + 2 x) + 2 x] e^{2 x} = C e^{2 x} .

Every $x$ term cancels — as it must, since $x e^{2 x}$ was chosen precisely to survive. So $C = 1$ and

y = A e^{x} + B e^{2 x} + x e^{2 x} .

✎ Worked example — "find

k

given the form"

Evaluate $k$ given that $k x \cos 2 x$ is a particular integral of $\frac{d^{2} y}{d x^{2}} + 4 y = \sin 2 x$ .

(The form has an $x$ in it because the CF is $A \cos 2 x + B \sin 2 x$ — this is the resonant case.) Differentiating twice by the product rule:

y = k x \cos 2 x, y^{'} = k \cos 2 x - 2 k x \sin 2 x, y^{''} = - 4 k \sin 2 x - 4 k x \cos 2 x .

Then the $x$ terms cancel:

y^{''} + 4 y = - 4 k \sin 2 x - 4 k x \cos 2 x + 4 k x \cos 2 x = - 4 k \sin 2 x .

Comparing with $\sin 2 x$ gives $- 4 k = 1$ , so $k = - \frac{1}{4}$ , and a full general solution would be $y = A \cos 2 x + B \sin 2 x - \frac{1}{4} x \cos 2 x$ .

§5 Using a given substitution

◆

x = e^{t}

turns variable coefficients into constant ones

For equations of the form $a x^{2} \frac{d^{2} y}{d x^{2}} + b x \frac{d y}{d x} + c y = f (x)$ , put $x = e^{t}$ . Then

x \frac{d y}{d x} = \frac{d y}{d t}, x^{2} \frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d t^{2}} - \frac{d y}{d t} .

Substituting leaves a constant-coefficient equation in $t$ , which you solve as in $§ 3$ and $§ 4$ — then convert back with $t = \ln x$ .

Deriving those two results

With $x = e^{t}$ we have $\frac{d x}{d t} = e^{t} = x$ , so by the chain rule $\frac{d y}{d t} = \frac{d y}{d x} \cdot x$ , which is the first result. Differentiating that with respect to $t$ :

\frac{d^{2} y}{d t^{2}} = \frac{d}{d t} (x \frac{d y}{d x}) = x \frac{d}{d x} (x \frac{d y}{d x}) = x (\frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}}) = \frac{d y}{d t} + x^{2} \frac{d^{2} y}{d x^{2}},

which rearranges to the second. $■$

✎ Worked example —

x = e^{t}

Use $x = e^{t}$ to solve $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - 4 y = 0$ for $x > 0$ .

(\frac{d^{2} y}{d t^{2}} - \frac{d y}{d t}) + \frac{d y}{d t} - 4 y = 0 ⟹ \frac{d^{2} y}{d t^{2}} - 4 y = 0 .

The first-derivative terms cancel. The auxiliary equation $m^{2} - 4 = 0$ gives $m = \pm 2$ , so $y = A e^{2 t} + B e^{- 2 t}$ . Converting back with $e^{t} = x$ :

y = A x^{2} + \frac{B}{x^{2}} .

Check $y = x^{2}$ : $x^{2} (2) + x (2 x) - 4 x^{2} = 2 x^{2} + 2 x^{2} - 4 x^{2} = 0$ ✓ Check $y = x^{- 2}$ : $x^{2} (6 x^{- 4}) + x (- 2 x^{- 3}) - 4 x^{- 2} = (6 - 2 - 4) x^{- 2} = 0$ ✓

✎ Worked example —

y = u x

makes it separable

Use $y = u x$ to solve $\frac{d y}{d x} = \frac{x + y}{x - y}$ .

With $y = u x$ , the product rule gives $\frac{d y}{d x} = u + x \frac{d u}{d x}$ , while the right-hand side becomes $\frac{x + u x}{x - u x} = \frac{1 + u}{1 - u}$ — every $x$ cancels, which is the whole point of the substitution. So

x \frac{d u}{d x} = \frac{1 + u}{1 - u} - u = \frac{(1 + u) - u (1 - u)}{1 - u} = \frac{1 + u^{2}}{1 - u} .

Now separate and split the left-hand integrand into two standard pieces:

\int \frac{1 - u}{1 + u^{2}} d u = \int \frac{d x}{x} ⟹ \tan^{- 1} u - \frac{1}{2} \ln (1 + u^{2}) = \ln | x | + A .

Finally put $u = \frac{y}{x}$ . Leaving the answer in this implicit form is perfectly acceptable — do not waste time trying to make $y$ the subject.

§6 Initial conditions and interpreting the model

◆ Find the general solution first — always

Apply the conditions only after you have the complete general solution including the PI. Applying them to the CF alone is a guaranteed loss of every remaining mark, because the constants come out wrong.

For a second order equation you will be given two conditions — typically $y$ and $\frac{d y}{d x}$ at one value of $x$ — giving two simultaneous equations in $A$ and $B$ .

✎ Worked example — with interpretation

Solve $y^{''} + 2 y^{'} + 5 y = 0$ given $y = 1$ and $y^{'} = 1$ when $x = 0$ , and describe the behaviour for large $x$ .

From $§ 3$ the general solution is $y = e^{- x} (A \cos 2 x + B \sin 2 x)$ . At $x = 0$ : $y = A = 1$ . Differentiating,

y^{'} = - e^{- x} (A \cos 2 x + B \sin 2 x) + e^{- x} (- 2 A \sin 2 x + 2 B \cos 2 x),

so at $x = 0$ , $y^{'} = - A + 2 B = 1$ . With $A = 1$ this gives $B = 1$ , and

y = e^{- x} (\cos 2 x + \sin 2 x) .

Interpretation. The bracket oscillates between $\pm \sqrt{2}$ forever, but the factor $e^{- x} \to 0$ . So the solution is a damped oscillation: it crosses zero repeatedly, with the size of each swing shrinking, and $y \to 0$ as $x \to \infty$ . In a modelling context that is a system returning to equilibrium while overshooting on the way.

§7 Interactive: the solution family, and a live check

Move the arbitrary constants and watch the whole family of solutions sweep out — every one of them satisfies the equation. The panel substitutes the stated solution back into the differential equation numerically and reports the largest residual over the plotted range: if that is essentially zero, the solution really is a solution.

Constant

A

1.0

Constant

B

1.0

your solution other members of the family the particular integral alone

every curve solves it ✎

§8 Examiner traps & A* checklist

⚠ The nine most common mark-losers

Not dividing through so the coefficient of $\frac{d y}{d x}$ is $1$ before finding the integrating factor.
Giving only the PI, or only the CF, as "the solution". The general solution is both.
Trying $C e^{b x}$ when $e^{b x}$ is already in the CF — multiply by $x$ .
Using only $C \cos p x$ when the right-hand side is $a \sin p x$ . You need both $\cos$ and $\sin$ in the trial form.
Applying the initial conditions to the CF before adding the PI.
Writing the repeated-root CF as $A e^{m x} + B e^{m x}$ — that collapses to one constant. It must be $(A + B x) e^{m x}$ .
Mishandling complex roots: the answer is $e^{p x} (A \cos q x + B \sin q x)$ , with $p$ the real part and $q$ the modulus of the imaginary part.
Forgetting the constant at the final integration of $I y = \int I Q d x$ .
Not answering the "interpret" part. If the question mentions a model, say in words what happens for large $x$ .

★ Before you turn the page

Does my answer have exactly the right number of arbitrary constants?
Have I substituted my PI back into the equation to confirm it works?
Does the solution satisfy the initial conditions when I put the numbers in?

§9 Video explainers (curated)

▶Integrating factor
Curated search — standard form, finding I, worked examples.↗ ▶CF + PI for second order
Curated search — all three root cases and the trial forms.↗ ▶Reduction by substitution
Curated search — x = eᵗ and y = ux substitutions.↗

§10 Check yourself

Score 0 / 5

Q1 · integrating factor

The integrating factor for $\frac{d y}{d x} + \frac{2}{x} y = x$ is:

Solution

$I = e^{\int P d x} = e^{\int (2 / x) d x} = e^{2 \ln x} = e^{\ln x^{2}} = x^{2}$ . The trap is stopping at $\int P d x = 2 \ln x$ — you must still exponentiate. Check: $x^{2} y^{'} + 2 x y = \frac{d}{d x} (x^{2} y)$ ✓

Q2 · repeated root

The complementary function of $\frac{d^{2} y}{d x^{2}} - 6 \frac{d y}{d x} + 9 y = 0$ is:

Solution

$m^{2} - 6 m + 9 = (m - 3)^{2} = 0$ gives the repeated root $m = 3$ , so the CF is $(A + B x) e^{3 x}$ . The third option is really $A e^{3 x} + B e^{3 x} = (A + B) e^{3 x}$ — one constant in disguise, so it cannot be a general solution of a second order equation.

Q3 · resonance

For $y^{''} - 4 y^{'} + 3 y = e^{3 x}$ , a suitable form for the particular integral is:

Solution

The auxiliary equation is $(m - 1) (m - 3) = 0$ , so the CF is $A e^{x} + B e^{3 x}$ and $e^{3 x}$ is already in it. Trying $C e^{3 x}$ would give $0 = e^{3 x}$ ; multiplying by $x$ once is enough, because the root $3$ is not repeated. (You would need $C x^{2} e^{3 x}$ only for a repeated root.)

Q4 · finding a given coefficient

Given that $k x \cos 2 x$ is a particular integral of $\frac{d^{2} y}{d x^{2}} + 4 y = \sin 2 x$ , the value of $k$ is:

Solution

With $y = k x \cos 2 x$ : $y^{'} = k \cos 2 x - 2 k x \sin 2 x$ and $y^{''} = - 4 k \sin 2 x - 4 k x \cos 2 x$ . The $x \cos 2 x$ terms cancel in $y^{''} + 4 y$ , leaving $- 4 k \sin 2 x$ . Comparing with $\sin 2 x$ gives $- 4 k = 1$ , so $k = - \frac{1}{4}$ . The sign is the whole question here.

Q5 · full method — try it on paper first

Solve $\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + 2 y = e^{3 x}$ , given that $y = 1$ and $\frac{d y}{d x} = 0$ when $x = 0$ . Describe the behaviour of $y$ for large $x$ .

Solution

Complementary function. The auxiliary equation is $m^{2} - 3 m + 2 = 0$ , i.e. $(m - 1) (m - 2) = 0$ , so $m = 1$ or $m = 2$ and

CF = A e^{x} + B e^{2 x} .

Particular integral. The right-hand side is $e^{3 x}$ , and $3$ is not a root of the auxiliary equation, so try $y = C e^{3 x}$ . Then $y^{'} = 3 C e^{3 x}$ , $y^{''} = 9 C e^{3 x}$ , and

9 C e^{3 x} - 9 C e^{3 x} + 2 C e^{3 x} = e^{3 x} ⟹ 2 C = 1 ⟹ C = \frac{1}{2} .

General solution. $y = A e^{x} + B e^{2 x} + \frac{1}{2} e^{3 x}$ .

Apply the conditions — to the full general solution, not just the CF. At $x = 0$ :

y = A + B + \frac{1}{2} = 1 ⟹ A + B = \frac{1}{2} .

Differentiating, $y^{'} = A e^{x} + 2 B e^{2 x} + \frac{3}{2} e^{3 x}$ , so at $x = 0$ :

A + 2 B + \frac{3}{2} = 0 ⟹ A + 2 B = - \frac{3}{2} .

Subtracting the first from the second gives $B = - 2$ , and then $A = \frac{1}{2} - (- 2) = \frac{5}{2}$ . Hence

y = \frac{5}{2} e^{x} - 2 e^{2 x} + \frac{1}{2} e^{3 x} .

Check: at $x = 0$ , $y = 2.5 - 2 + 0.5 = 1$ ✓ and $y^{'} = 2.5 - 4 + 1.5 = 0$ ✓

Behaviour for large $x$ . All three terms grow, but $e^{3 x}$ grows fastest, so it dominates: $y \approx \frac{1}{2} e^{3 x} \to + \infty$ . The solution grows without bound, at the rate set by the particular integral rather than by the complementary function. $■$

FP2 · Topic 13 · Differential Equations