Further Pure — Differential equations: A-level Further Maths OCR revision notes

4.10

First-order differential equations

Linear first-order equations: the integrating factor 4.10

Definitions

Linear first-order equation: one that can be written $\frac{d y}{d x} + P (x) y = Q (x)$ .
Integrating factor: $I (x) = e^{\int P (x) d x}$ , chosen so that $I \frac{d y}{d x} + I P y = \frac{d}{d x} (I y)$ .
General solution: the family of all solutions, containing an arbitrary constant. A particular solution is the member fixed by a boundary condition.

Key results

Multiplying $\frac{d y}{d x} + P y = Q$ by $I$ gives $\frac{d}{d x} (I y) = I Q$ , so $I y = \int I Q d x$ .
This works because $I'=PI$ , so the product rule gives $(Iy)'=Iy'+PIy$ exactly.

Method

Divide through so the coefficient of $\frac{d y}{d x}$ is 1, and read off $P (x)$ .
Find $I = e^{\int P d x}$ and simplify it: $e^{2 l n x} = x^{2}$ , $e^{- l n c o s x} = sec x$ .
Write $\frac{d}{d x} (I y) = I Q$ , integrate, and include the constant before dividing by $I$ .
Apply the boundary condition last, to the whole solution.

In practiceSolve $x \frac{d y}{d x} + 2 y = x^{2}$ for $x > 0$ , given $y = 1$ when $x = 1$ .

$\frac{d y}{d x} + \frac{2}{x} y = x$ divide by $x$ , so $P = \frac{2}{x}$
$I = e^{\int \frac{2}{x} d x} = e^{2 l n x} = x^{2}$
$\frac{d}{d x} (x^{2} y) = x^{3} ⟹ x^{2} y = \frac{1}{4} x^{4} + c$ add the constant before dividing
$y = \frac{1}{4} x^{2} + \frac{c}{x ^{2}}, y (1) = 1 ⟹ c = \frac{3}{4}$
$y = \frac{1}{4} x^{2} + \frac{3}{4 x ^{2}}$ check: $xy'+2y=x^2$

Notes

No constant is needed in $\int P d x$ : it would only multiply $I$ by a constant that cancels.
The constant of integration must be added before dividing by $I$ . Writing $y = \frac{1}{I} \int I Q d x$ and adding $+ C$ at the end loses the term $C / I$ , which is the whole family of curves on the sheet.
If the equation is also separable, either method works. If it is neither, the question will give a substitution that turns it into one of the two.

Choosing a method and using substitutions 4.10

Key results

Separable: $\frac{d y}{d x} = f (x) g (y)$ gives $\int \frac{d y}{g ( y )} = \int f (x) d x$ .
Linear: $\frac{d y}{d x} + P (x) y = Q (x)$ , solved with an integrating factor.
A given substitution, such as $y = vx$ or $z = y^{- 1}$ , changes the dependent variable so that the new equation is separable or linear.

Method

For a substitution $y = vx$ : differentiate with the product rule, $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , then replace both $y$ and $\frac{d y}{d x}$ so that only $v$ and $x$ remain.
Solve for $v$ , then substitute back to give $y$ in terms of $x$ .

In practiceUse $y = vx$ to solve $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}$ for $x > 0$ .

$\frac{d y}{d x} = v + x \frac{d v}{d x}$ product rule
$v + x \frac{d v}{d x} = \frac{x ^{2} + v ^{2} x ^{2}}{v x ^{2}} = \frac{1}{v} + v$ replace $y$ and $\frac{d y}{d x}$
$x \frac{d v}{d x} = \frac{1}{v} ⟹ \int v d v = \int \frac{d x}{x}$ now separable
$\frac{1}{2} v^{2} = ln x + c ⟹ y^{2} = 2 x^{2} (ln x + c)$ substitute back $v = \frac{y}{x}$

Notes

When separating, dividing by $g (y)$ can lose a constant solution $y = c$ with $g (c) = 0$ . Check whether it is needed.
Always state the final answer in the requested form: explicit $y = f (x)$ if asked for, not an implicit relation.

Worked examples

Worked example

Solve:

x \frac{d y}{d x} + 2 y = x^{2}

for $x > 0$ , given that $y = 1$ when $x = 1$ .

Show worked solution

Divide by $x$ :

\frac{d y}{d x} + \frac{2}{x} y = x

so:

P = \frac{2}{x}

and $I = e^{2 l n x} = x^{2}$ .

Then:

\frac{d}{d x} (x^{2} y) = x^{3}

so:

x^{2} y = \frac{x ^{4}}{4} + C

At $x = 1$ :

1 = \frac{1}{4} + C

C = \frac{3}{4}

Hence:

y = \frac{x ^{2}}{4} + \frac{3}{4 x ^{2}}

Worked example

Solve:

\frac{d y}{d x} + 2 y = e^{x}

given that $y = 1$ when $x = 0$ .

Show worked solution

$I = e^{2 x}$ , so:

\frac{d}{d x} (e^{2 x} y) = e^{3 x}

and:

e^{2 x} y = \frac{1}{3} e^{3 x} + C

giving:

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

From $y (0) = 1$ ,

C = \frac{2}{3}

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

4.10

Second-order linear differential equations

Homogeneous second-order equations 4.10

Definitions

Homogeneous: $a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0$ , with constants $a$ , $b$ , $c$ .
Auxiliary equation: $a m^{2} + bm + c = 0$ , found by trying $y = e^{m x}$ .

Key results

Distinct real roots $m_{1}$ , $m_{2}$ : $y = A e^{m_{1} x} + B e^{m_{2} x}$ .
Repeated root $m$ : $y = (A + B x) e^{m x}$ .
Complex roots $p \pm q i$ : $y = e^{p x} (A cos q x + B sin q x)$ .
Purely imaginary roots $\pm q i$ : $y = A cos q x + B sin q x$ , the special case $p = 0$ .

Method

Write down the auxiliary equation and compute $b^{2} - 4 a c$ to identify the case.
Write the general solution with two arbitrary constants.
Use two conditions, usually $y$ and $\frac{d y}{d x}$ at $x = 0$ , to find $A$ and $B$ .

In practiceSolve $y''+4y'+13y=0$ with $y (0) = 2$ and $y'(0)=0$ .

$m^{2} + 4 m + 13 = 0 ⟹ m = - 2 \pm 3 i$ $b^{2} - 4 a c = - 36 < 0$ : complex roots
$y = e^{- 2 x} (A cos 3 x + B sin 3 x)$
$y (0) = A = 2$
$\displaystyle y'(0)=-2A+3B=0\implies B=\tfrac43$ differentiate with the product rule, then put $x = 0$
$y = e^{- 2 x} (2 cos 3 x + \frac{4}{3} sin 3 x)$

Notes

A second-order equation needs exactly two arbitrary constants. A repeated-root answer $A e^{m x}$ alone is incomplete.
The complex case follows from $e^{(p \pm q i) x} = e^{p x} (cos q x \pm i sin q x)$ ; real combinations of the two give the cosine and sine terms.
In the sheet's example the sign of $k^{2} - 4$ decides the behaviour: $k < 2$ oscillates, $k = 2$ is critical, $k > 2$ creeps back.

Non-homogeneous equations: complementary function and particular integral 4.10

Definitions

Complementary function (CF): the general solution of the related homogeneous equation.
Particular integral (PI): any one solution of the full equation $ay''+by'+cy=f(x)$ .

Key results

General solution = CF + PI. The CF carries both arbitrary constants; the PI has none.
Standard trial functions: for $f (x)$ a polynomial of degree $n$ , a general polynomial of degree $n$ ; for $k e^{p x}$ , $λ e^{p x}$ ; for $k cos ω x$ or $k sin ω x$ , $λ cos ω x + μ sin ω x$ .

Method

Find the CF first, because it decides the trial function.
If the trial function, or part of it, already appears in the CF, multiply it by $x$ . If that also appears (a repeated root), multiply by $x^{2}$ .
Substitute the trial function into the full equation and compare coefficients.
Add CF and PI, then apply the initial conditions to the complete solution.

In practiceFind the general solution of $y''-3y'+2y=4e^{2x}$ .

$m^{2} - 3 m + 2 = 0 ⟹ m = 1, 2 : CF y = A e^{x} + B e^{2 x}$ CF first
$trial y = λ x e^{2 x}$ $e^{2 x}$ is already in the CF, so multiply by $x$
$\displaystyle y'=\lambda e^{2x}(1+2x),\qquad y''=\lambda e^{2x}(4+4x)$
$λ e^{2 x} [(4 + 4 x) - 3 (1 + 2 x) + 2 x] = λ e^{2 x} = 4 e^{2 x} ⟹ λ = 4$ the $x$ terms cancel, as they must
$y = A e^{x} + B e^{2 x} + 4 x e^{2 x}$ CF + PI

Notes

Applying initial conditions to the CF before adding the PI is a common and costly error.
For a trigonometric $f (x)$ include both $cos$ and $sin$ in the trial, even if $f (x)$ has only one. The $y'$ term mixes them.

Worked examples

Worked example

Find the general solution of:

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

Show worked solution

CF: $m^{2} - 3 m + 2 = 0$ gives $m = 1, 2$ , so:

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

PI: try $y = λ x + μ$ ; then:

- 3 λ + 2 (λ x + μ) = 4 x

so $λ = 2$ and $2 μ - 6 = 0$ , $μ = 3$ .

General solution:

y = A e^{x} + B e^{2 x} + 2 x + 3

Worked example

Find the general solution of:

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 3 e^{2 x}

Show worked solution

CF:

m^{2} - m - 2 = (m - 2) (m + 1) = 0

so:

y_{CF} = A e^{2 x} + B e^{- x}

The trial $λ e^{2 x}$ is in the CF, so try $λ x e^{2 x}$ :

y&#039;=\lambda e^{2x}(1+2x)

y&#039;&#039;=\lambda e^{2x}(4+4x)

Substituting,

λ e^{2 x} [(4 + 4 x) - (1 + 2 x) - 2 x] = 3 λ e^{2 x} = 3 e^{2 x}

so $λ = 1$ .

General solution:

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

4.10

Simple harmonic motion and damping

Simple harmonic motion 4.10

Definitions

Simple harmonic motion (SHM): motion in which the acceleration is proportional to the displacement from a fixed point and directed towards it, $\overset{x}{¨} = - ω^{2} x$ .

Key results

General solution: $x = A cos ω t + B sin ω t$ , equivalently $x = a sin (ω t + φ)$ with amplitude $a = A^{2} + B^{2}$ .
Period $T = \frac{2 π}{ω}$ , independent of the amplitude.
Speed: $v^{2} = ω^{2} (a^{2} - x^{2})$ , so the maximum speed is $aω$ at $x = 0$ .
Maximum acceleration $a ω^{2}$ , at $x = \pm a$ .

Method

Show a motion is SHM by deriving an equation of the form $\overset{x}{¨} = - ω^{2} x$ from Newton's second law, measuring $x$ from the equilibrium position.
Use $v^{2} = ω^{2} (a^{2} - x^{2})$ for speed questions and the explicit solution for timing questions.

In practiceA particle of mass $0.5$ kg on a spring of stiffness $8$ N m $^{- 1}$ is released from rest $0.1$ m from equilibrium. Find the period and the greatest speed.

$0.5 \overset{x}{¨} = - 8 x ⟹ \overset{x}{¨} = - 16 x$ Newton's second law, $x$ measured from equilibrium
$ω = 4, T = \frac{2 π}{4} = \frac{π}{2} \approx 1.57 s$ the form $\overset{x}{¨} = - ω^{2} x$ : SHM
$v_{m a x} = aω = 0.1 \times 4 = 0.4 m s^{- 1}$ at the equilibrium position

Notes

$v^{2} = ω^{2} (a^{2} - x^{2})$ follows from writing $\overset{x}{¨} = v \frac{d v}{d x}$ and integrating, a useful derivation to know.
Set the calculator to radians for any question involving $ω t$ .

Damped and forced oscillations 4.10

Definitions

Damping: a resistance proportional to velocity, giving $\overset{x}{¨} + 2 k \overset{x}{˙} + ω^{2} x = 0$ with $k > 0$ .
Forcing: an external driving term on the right, $\overset{x}{¨} + 2 k \overset{x}{˙} + ω^{2} x = f (t)$ .

Key results

Heavy damping ( $k > ω$ ): distinct real negative roots; no oscillation.
Critical damping ( $k = ω$ ): a repeated root; the fastest return without overshoot.
Light damping ( $k < ω$ ): complex roots; oscillation inside a decaying exponential envelope.
With damping, the CF decays to zero, so in the long term the motion is the PI: the steady state.

Method

Form the equation from Newton's second law, with resistance and restoring forces opposite to velocity and displacement.
Divide by the mass, classify with the discriminant, then solve as for any second-order equation.

In practiceClassify the motion $\overset{x}{¨} + 6 \overset{x}{˙} + ω^{2} x = 0$ when $ω^{2} = 5$ , $9$ and $25$ .

$m^{2} + 6 m + ω^{2} = 0, Δ = 36 - 4 ω^{2}$ auxiliary equation
$ω^{2} = 5 : Δ = 16 > 0, m = - 1, - 5$ heavy damping: $x = A e^{- t} + B e^{- 5 t}$
$ω^{2} = 9 : Δ = 0, m = - 3$ critical damping: $x = (A + Bt) e^{- 3 t}$
$ω^{2} = 25 : Δ = - 64 < 0, m = - 3 \pm 4 i$ light damping: $x = e^{- 3 t} (A cos 4 t + B sin 4 t)$

Notes

With $k > 0$ and $ω > 0$ every CF term decays, whichever case applies; a growing solution signals a sign error in the model.
When asked about long-term behaviour, say which terms vanish and why, then describe the PI.
Resonance: with no damping, forcing at the natural frequency, $\overset{x}{¨} + ω^{2} x = F cos ω t$ , makes the usual trial fail because it is already in the CF. The trial $t (λ cos ω t + μ sin ω t)$ gives the PI $x = \frac{F}{2 ω} t sin ω t$ : an oscillation whose amplitude grows without bound.

Worked examples

Worked example

A system satisfies:

\overset{x}{¨} + 2 \overset{x}{˙} + 5 x = 5

starting at rest with $x = 0$ .

Find $x$ in terms of $t$ and describe the long-term motion.

Show worked solution

CF: $m^{2} + 2 m + 5 = 0$ gives $m = - 1 \pm 2 i$ , so:

x_{CF} = e^{- t} (A cos 2 t + B sin 2 t)

PI: $x = 1$ .

So:

x = 1 + e^{- t} (A cos 2 t + B sin 2 t)

$x (0) = 0$ gives $A = - 1$ .

\overset{x}{˙} (0) = - A + 2 B = 0

gives:

B = - \frac{1}{2}

Hence:

x = 1 - e^{- t} (cos 2 t + \frac{1}{2} sin 2 t)

As $t \to \infty$ , $e^{- t} \to 0$ , so the oscillation dies away and $x \to 1$ , as on the sheet.

Worked example

A particle moves with SHM of amplitude 0.5 m and period $π$ s.

Find its greatest speed and its speed when 0.3 m from the centre.

Show worked solution

ω = \frac{2 π}{T} = 2

Greatest speed:

aω = 0.5 \times 2 = 1 m s^{- 1}

At $x = 0.3$ :

v^{2} = 4 (0.25 - 0.09) = 0.64

so $v = 0.8 m s^{- 1}$ .

4.10

Systems of first-order equations

Coupled first-order systems 4.10

Definitions

Coupled system: two first-order equations in which the rate of change of each variable depends on both, such as $\overset{x}{˙} = a x + b y$ , $\overset{y}{˙} = c x + d y$ .

Method

Rearrange one equation for the variable to be eliminated, here $y = \frac{x ˙ - a x}{b}$ .
Differentiate the first equation: from $\overset{x}{˙} = a x + b y$ , $\overset{x}{¨} = a \overset{x}{˙} + b \overset{y}{˙}$ .
Substitute for $\overset{y}{˙}$ and then $y$ to obtain a second-order equation in $x$ alone, and solve it.
Find $y$ from the rearranged equation, not by solving a second differential equation, so no extra constants appear.

In practiceSolve $\overset{x}{˙} = x + 2 y$ , $\overset{y}{˙} = 2 x + y$ .

$y = \frac{1}{2} (\overset{x}{˙} - x)$ rearrange the first equation
$\overset{x}{¨} = \overset{x}{˙} + 2 \overset{y}{˙} = \overset{x}{˙} + 2 (2 x + y) = \overset{x}{˙} + 4 x + (\overset{x}{˙} - x)$ differentiate, then substitute
$\overset{x}{¨} - 2 \overset{x}{˙} - 3 x = 0 ⟹ m = 3, - 1 : x = A e^{3 t} + B e^{- t}$
$y = \frac{1}{2} (\overset{x}{˙} - x) = A e^{3 t} - B e^{- t}$ no new constants; check: $\overset{y}{˙} = 2 x + y$

Notes

Coupled systems model predator and prey, competing species, mixing tanks and connected circuits.
The final answer has exactly two arbitrary constants, shared between $x$ and $y$ .

Worked example

Solve $\overset{x}{˙} = 4 x - 2 y$ , $\overset{y}{˙} = 3 x - y$ , given that $x = 3$ and $y = 4$ when $t = 0$ .

Show worked solution

From the first equation $2 y = 4 x - \overset{x}{˙}$ .

Differentiating it:

\overset{x}{¨} = 4 \overset{x}{˙} - 2 \overset{y}{˙} = 4 \overset{x}{˙} - 6 x + 2 y = 4 \overset{x}{˙} - 6 x + 4 x - \overset{x}{˙}

so:

\overset{x}{¨} - 3 \overset{x}{˙} + 2 x = 0

and:

x = A e^{t} + B e^{2 t}

Then:

y = \frac{4 x - x ˙}{2} = \frac{3}{2} A e^{t} + B e^{2 t}

From $t = 0$ : $A + B = 3$ and:

\frac{3}{2} A + B = 4

so $A = 2$ , $B = 1$ .

Hence $x = 2 e^{t} + e^{2 t}$ , $y = 3 e^{t} + e^{2 t}$ .