Differential Equations

Differential Equations

Differential equations are equations involving derivatives of an unknown function. They arise naturally whenever a rate of change is related to the current state of a system — from population growth to electrical circuits. This chapter covers first-order equations (separable and integrating factor methods), second-order linear equations with constant coefficients, and modelling applications.

Slope Field Explorer

Adjust the parameters in the graph above to explore the relationships between variables.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	First-order (separable, integrating factor), growth/decay models
Edexcel	FP2	First-order + second-order linear with constant coefficients
OCR (A)	Paper 1	First-order + second-order linear with constant coefficients
CIE	P2	First-order (separable, integrating factor); second-order in P2

All boards examine first-order ODEs. CIE and Edexcel require second-order linear ODEs. AQA

focuses on first-order equations with growth and decay modelling. The formula booklet gives the integrating factor formula on Edexcel; AQA and OCR students must know it. :::

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1. First-Order ODEs: Separable Equations

1.1 Definition

Definition. A first-order ordinary differential equation (ODE) is separable if it can be written in the form

$\frac{dy}{dx} = f(x)\,g(y)$

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone.

1.2 Method

Separate the variables and integrate both sides:

$\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$

$\boxed{\int \frac{1}{g(y)}\,dy = \int f(x)\,dx + C}$

1.3 Worked example

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 2$.

Separate: $y\,dy = x\,dx$.

Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$.

Using $y(0) = 2$: $\dfrac{4}{2} = 0 + C \implies C = 2$.

$y^2 = x^2 + 4, \quad y = \sqrt{x^2 + 4}$

(We take the positive root since $y(0) = 2 > 0$.)

1.4 Domain restrictions

When dividing by $g(y)$ during separation, we implicitly assume $g(y) \neq 0$. If

$g(y_0) = 0$, then $y = y_0$ is a constant (equilibrium) solution that may not appear in the general solution. Always check for these. :::

Example. $\dfrac{dy}{dx} = y(1-y)$.

Separating: $\displaystyle\int\frac{1}{y(1-y)}\,dy = \int dx$.

Partial fractions: $\dfrac{1}{y(1-y)} = \dfrac{1}{y} + \dfrac{1}{1-y}$.

$\ln|y| - \ln|1-y| = x + C$.

$\ln\left|\dfrac{y}{1-y}\right| = x + C$.

$\dfrac{y}{1-y} = Ae^x$ where $A = \pm e^C$.

$\boxed{y = \frac{Ae^x}{1 + Ae^x}}$

But note $y = 0$ and $y = 1$ are also solutions (equilibrium solutions), corresponding to $A = 0$ and the limiting case $A \to \infty$.

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2. First-Order ODEs: Integrating Factor Method

2.1 Standard form

Definition. A first-order linear ODE has the form

$\frac{dy}{dx} + P(x)\,y = Q(x)$

where $P(x)$ and $Q(x)$ are continuous functions of $x$.

2.2 The integrating factor

The integrating factor is

$\boxed{\mu(x) = e^{\int P(x)\,dx}}$

2.3 Derivation of the method

Proof of the integrating factor technique

Multiply the ODE $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ by $\mu(x)$:

$\mu\frac{dy}{dx} + \mu P\,y = \mu Q$

The left-hand side is the derivative of $\mu y$ because:

$\frac{d}{dx}(\mu y) = \mu\frac{dy}{dx} + y\frac◆LB◆d\mu◆RB◆◆LB◆dx◆RB◆$

Since $\dfrac◆LB◆d\mu◆RB◆◆LB◆dx◆RB◆ = \mu \cdot P(x)$ (because $\mu = e^{\int P\,dx}$, so $\mu' = P \cdot e^{\int P\,dx} = P\mu$):

$\frac{d}{dx}(\mu y) = \mu\frac{dy}{dx} + y \cdot P\mu = \mu\frac{dy}{dx} + \mu P\,y$

This is exactly the left-hand side. Therefore:

$\frac{d}{dx}(\mu y) = \mu Q$

Integrating both sides:

$\boxed{\mu y = \int \mu Q\,dx + C}$

$\boxed{y = \frac◆LB◆1◆RB◆◆LB◆\mu◆RB◆\left(\int \mu\, Q\,dx + C\right)}$

$\square$

2.4 Worked example

Solve $\dfrac{dy}{dx} + \dfrac{2}{x}\,y = x^2$ for $x > 0$.

Here $P(x) = \dfrac{2}{x}$, $Q(x) = x^2$.

$\mu = e^{\int 2/x\,dx} = e^{2\ln x} = x^2$

Multiply through: $x^2\dfrac{dy}{dx} + 2xy = x^4$.

Left-hand side is $\dfrac{d}{dx}(x^2 y) = x^4$.

Integrate: $x^2 y = \dfrac{x^5}{5} + C$.

$\boxed{y = \frac{x^3}{5} + \frac{C}{x^2}}$

2.5 Worked example with boundary condition

Solve $\dfrac{dy}{dx} - 3y = e^{2x}$, given $y(0) = 1$.

$P(x) = -3$, $Q(x) = e^{2x}$.

$\mu = e^{\int -3\,dx} = e^{-3x}$

$\frac{d}{dx}(e^{-3x}y) = e^{-3x} \cdot e^{2x} = e^{-x}$

$e^{-3x}y = \int e^{-x}\,dx = -e^{-x} + C$

$y = -e^{2x} + Ce^{3x}$

Using $y(0) = 1$: $1 = -1 + C \implies C = 2$.

$\boxed{y = 2e^{3x} - e^{2x}}$

The constant of integration in $\mu = e^{\int P\,dx}$ can be omitted (absorbed into $C$).

Always choose the simplest antiderivative. :::

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3. Second-Order Linear ODEs with Constant Coefficients

3.1 General form

Definition. A second-order linear ODE with constant coefficients has the form

$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$

where $a$, $b$, $c$ are constants with $a \neq 0$.

The equation is homogeneous when $f(x) = 0$:

$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$

3.2 The auxiliary equation

To solve the homogeneous equation, try $y = e^{mx}$. Then $y' = me^{mx}$ and $y'' = m^2 e^{mx}$.

Substituting: $am^2 e^{mx} + bme^{mx} + ce^{mx} = 0$.

Since $e^{mx} \neq 0$ for all $x$:

$\boxed{am^2 + bm + c = 0}$

This is the auxiliary equation (or characteristic equation).

3.3 Three cases

The nature of the roots of $am^2 + bm + c = 0$ determines the form of the general solution.

Case 1: Two distinct real roots $m_1 \neq m_2$ (discriminant $\Delta = b^2 - 4ac > 0$):

$\boxed{y = Ae^{m_1 x} + Be^{m_2 x}}$

Case 2: Repeated real root $m$ (discriminant $\Delta = 0$):

$\boxed{y = (A + Bx)e^{mx}}$

Case 3: Complex conjugate roots $m = \alpha \pm \beta i$ (discriminant $\Delta < 0$):

$\boxed{y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)}$

Proof of the general solution for Case 3 (complex roots)

When $m = \alpha + \beta i$ and $\overline{m} = \alpha - \beta i$, the two linearly independent solutions are $e^{(\alpha+\beta i)x}$ and $e^{(\alpha-\beta i)x}$.

Using Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$:

$e^{(\alpha+\beta i)x} = e^{\alpha x}(\cos\beta x + i\sin\beta x)$ $e^{(\alpha-\beta i)x} = e^{\alpha x}(\cos\beta x - i\sin\beta x)$

Any linear combination $C_1 e^{(\alpha+\beta i)x} + C_2 e^{(\alpha-\beta i)x}$ can be rewritten as:

$e^{\alpha x}\left[(C_1+C_2)\cos\beta x + i(C_1-C_2)\sin\beta x\right]$

Setting $A = C_1 + C_2$ and $B = i(C_1 - C_2)$ (which are real when $C_1$ and $C_2$ are complex conjugates), we obtain:

$y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)$

$\square$

3.4 Worked examples

Example (Case 1). Solve $y'' - 5y' + 6y = 0$.

Auxiliary: $m^2 - 5m + 6 = 0 \implies (m-2)(m-3) = 0 \implies m = 2, 3$.

$y = Ae^{2x} + Be^{3x}$

Example (Case 2). Solve $y'' - 4y' + 4y = 0$.

Auxiliary: $m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \implies m = 2$ (repeated).

$y = (A + Bx)e^{2x}$

Example (Case 3). Solve $y'' + 2y' + 5y = 0$.

Auxiliary: $m^2 + 2m + 5 = 0 \implies m = \dfrac◆LB◆-2 \pm \sqrt{4-20}◆RB◆◆LB◆2◆RB◆ = -1 \pm 2i$.

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

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4. Non-Homogeneous Second-Order ODEs: Particular Solutions

4.1 Structure of the general solution

Theorem (Superposition). If $y_h$ is the general solution of the homogeneous equation $a y'' + by' + cy = 0$ (the complementary function, CF), and $y_p$ is any particular solution of $a y'' + by' + cy = f(x)$ (a particular integral, PI), then the general solution is

$\boxed{y = y_h + y_p}$

Proof of the superposition principle

Let $y_1$ satisfy $a y_1'' + by_1' + cy_1 = 0$ and $y_p$ satisfy $a y_p'' + by_p' + cy_p = f(x)$.

Define $y = y_1 + y_p$. Then:

$ay'' + by' + cy = a(y_1'' + y_p'') + b(y_1' + y_p') + c(y_1 + y_p)$

$= (ay_1'' + by_1' + cy_1) + (ay_p'' + by_p' + cy_p) = 0 + f(x) = f(x)$

So $y_1 + y_p$ satisfies the non-homogeneous equation. Since $y_1$ contains two arbitrary constants, the general solution $y = y_h + y_p$ also contains two arbitrary constants. $\square$

4.2 Method of undetermined coefficients

To find $y_p$, guess the form based on $f(x)$, then determine coefficients by substitution.

$f(x)$	Trial $y_p$ (if not in CF)	If in CF, multiply by $x$
$k$ (constant)	$c$	$cx$ then $cx^2$
$kx + \ell$	$px + q$	$x(px+q)$ then $x^2(px+q)$
$ke^{\alpha x}$	$ce^{\alpha x}$	$cxe^{\alpha x}$
$k\cos\omega x$	$c\cos\omega x + d\sin\omega x$	$x(c\cos\omega x + d\sin\omega x)$
$k\sin\omega x$	$c\cos\omega x + d\sin\omega x$	$x(c\cos\omega x + d\sin\omega x)$
Polynomial	General polynomial of same degree	Multiply by $x$ as needed

If any term in your trial $y_p$ already appears in the complementary function $y_h$,

multiply the entire trial by $x$. If it still appears, multiply by $x^2$. :::

4.3 Worked examples

Example 1. Solve $y'' - 3y' + 2y = 4e^{3x}$.

CF: $m^2 - 3m + 2 = 0 \implies (m-1)(m-2) = 0 \implies m = 1, 2$. $y_h = Ae^x + Be^{2x}$.

PI: Since $f(x) = 4e^{3x}$ and $e^{3x}$ is not in the CF, try $y_p = ce^{3x}$. $y_p' = 3ce^{3x}$, $y_p'' = 9ce^{3x}$.

Substitute: $9ce^{3x} - 9ce^{3x} + 2ce^{3x} = 4e^{3x} \implies 2c = 4 \implies c = 2$.

$\boxed{y = Ae^x + Be^{2x} + 2e^{3x}}$

Example 2. Solve $y'' + y = \sin x$.

CF: $m^2 + 1 = 0 \implies m = \pm i$. $y_h = A\cos x + B\sin x$.

PI: $f(x) = \sin x$. The terms $\cos x$ and $\sin x$ are in the CF, so multiply by $x$: $y_p = x(c\cos x + d\sin x)$.

$y_p' = (c\cos x + d\sin x) + x(-c\sin x + d\cos x)$ $y_p'' = 2(-c\sin x + d\cos x) + x(-c\cos x - d\sin x)$

Substitute into $y'' + y = \sin x$:

$2(-c\sin x + d\cos x) + x(-c\cos x - d\sin x) + x(c\cos x + d\sin x) = \sin x$

$2(-c\sin x + d\cos x) = \sin x$

Comparing: $-2c = 1 \implies c = -\dfrac{1}{2}$, and $2d = 0 \implies d = 0$.

$\boxed{y = A\cos x + B\sin x - \frac{1}{2}x\cos x}$

Example 3. Solve $y'' - 4y' + 4y = 3x + 2$.

CF: $m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0$. $y_h = (A + Bx)e^{2x}$.

PI: Try $y_p = px + q$. $y_p' = p$, $y_p'' = 0$.

$0 - 4p + 4(px + q) = 3x + 2$. $4px + (4q - 4p) = 3x + 2$.

$4p = 3 \implies p = \dfrac{3}{4}$. $4q - 3 = 2 \implies q = \dfrac{5}{4}$.

$\boxed{y = (A + Bx)e^{2x} + \frac{3}{4}x + \frac{5}{4}}$

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5. Growth and Decay Models

5.1 Exponential growth and decay

The ODE $\dfrac{dy}{dt} = ky$ (where $k$ is a constant) has solution

$\boxed{y = y_0 e^{kt}}$

where $y_0 = y(0)$. If $k > 0$: exponential growth. If $k < 0$: exponential decay.

Proof. Separable: $\dfrac{1}{y}\dfrac{dy}{dt} = k \implies \displaystyle\int \frac{1}{y}\,dy = \int k\,dt \implies \ln y = kt + C$.

$y = e^{kt+C} = Ae^{kt}$. With $y(0) = y_0$: $A = y_0$. $\square$

5.2 Newton's law of cooling

Definition. Newton's law of cooling states that the rate of temperature change of a body is proportional to the difference between its temperature $T$ and the ambient temperature $T_a$:

$\frac{dT}{dt} = -k(T - T_a)$

where $k > 0$ is the cooling constant.

Solution. Let $\theta = T - T_a$. Then $\dfrac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ = -k\theta$, giving $\theta = \theta_0 e^{-kt}$.

$\boxed{T = T_a + (T_0 - T_a)e^{-kt}}$

Example. A cup of tea at $90°C$ is placed in a room at $20°C$. After 10 minutes the temperature is $60°C$. Find the temperature after 20 minutes.

$T_0 = 90$, $T_a = 20$. $T(10) = 60$:

$60 = 20 + 70e^{-10k} \implies e^{-10k} = \dfrac{40}{70} = \dfrac{4}{7} \implies k = \dfrac{1}{10}\ln\dfrac{7}{4}$.

$T(20) = 20 + 70e^{-20k} = 20 + 70\left(\dfrac{4}{7}\right)^2 = 20 + 70 \cdot \dfrac{16}{49} = 20 + \dfrac{160}{7} \approx 42.9°C$.

5.3 Population growth with carrying capacity

The logistic equation models population growth with a carrying capacity $M$:

$\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)$

This is separable. The solution is

$\boxed{P(t) = \frac{M}{1 + Ae^{-kt}}}$

where $A = \dfrac{M - P_0}{P_0}$.

5.4 Mixing problems

Example. A tank contains 100 litres of water with 10 kg of salt. Pure water flows in at 5 L/min, and the mixture flows out at 5 L/min. Find the amount of salt after $t$ minutes.

Let $S(t)$ be the amount of salt (kg). Rate in = 0. Rate out = concentration $\times$ flow rate $= \dfrac{S}{100} \times 5 = \dfrac{S}{20}$.

$\frac{dS}{dt} = -\frac{S}{20}$

$S = 10e^{-t/20}$.

After 20 minutes: $S(20) = 10e^{-1} \approx 3.68$ kg.

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6. Proof of the General Solution of the Second-Order Homogeneous ODE

Proof that the CF gives the general solution

We must show that for all three cases, the proposed general solution has two arbitrary constants and satisfies the ODE.

Case 1 ($m_1 \neq m_2$, real). Let $y_1 = e^{m_1 x}$ and $y_2 = e^{m_2 x}$.

Substituting $y_1$ into $ay'' + by' + cy = 0$: $am_1^2 e^{m_1 x} + bm_1 e^{m_1 x} + ce^{m_1 x} = (am_1^2 + bm_1 + c)e^{m_1 x} = 0$ since $m_1$ is a root. Similarly for $y_2$.

The Wronskian is $W = y_1 y_2' - y_1' y_2 = m_2 e^{(m_1+m_2)x} - m_1 e^{(m_1+m_2)x} = (m_2 - m_1)e^{(m_1+m_2)x} \neq 0$.

Since $W \neq 0$, $y_1$ and $y_2$ are linearly independent, so $y = Ae^{m_1 x} + Be^{m_2 x}$ is the general solution.

Case 2 ($m$ repeated). $y_1 = e^{mx}$ is one solution. We need a second linearly independent solution. Try $y_2 = xe^{mx}$.

$y_2' = e^{mx}(1 + mx)$, $y_2'' = e^{mx}(2m + m^2 x)$.

$ay_2'' + by_2' + cy_2 = e^{mx}\left[a(2m + m^2 x) + b(1 + mx) + cx\right]$

$= e^{mx}\left[(am^2 + bm + c)x + (2am + b)\right]$

The coefficient of $x$ is zero since $m$ satisfies the auxiliary equation. The constant: $2am + b = 0$ when the root is repeated (since $m = -b/2a$). So $y_2$ is also a solution.

The Wronskian: $W = e^{mx} \cdot e^{mx}(1+mx) - me^{mx} \cdot xe^{mx} = e^{2mx} \neq 0$.

Therefore $y = (A + Bx)e^{mx}$ is the general solution.

$\square$

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Problems

Problem 1

Solve $\dfrac{dy}{dx} = \dfrac{3x^2}{y^2}$ given $y(1) = 2$.

Solution 1

Separate: $y^2\,dy = 3x^2\,dx$.

Integrate: $\dfrac{y^3}{3} = x^3 + C$.

$y(1) = 2$: $\dfrac{8}{3} = 1 + C \implies C = \dfrac{5}{3}$.

$y^3 = 3x^3 + 5 \implies \boxed{y = \sqrt[3]{3x^3 + 5}}$.

If you get this wrong, revise: Separable Equations — Section 1.

Problem 2

Solve $\dfrac{dy}{dx} + \dfrac{3}{x}\,y = x^2$ for $x > 0$, given $y(1) = 0$.

Solution 2

$P(x) = 3/x$, $Q(x) = x^2$.

$\mu = e^{\int 3/x\,dx} = e^{3\ln x} = x^3$.

$\dfrac{d}{dx}(x^3 y) = x^3 \cdot x^2 = x^5$.

$x^3 y = \dfrac{x^6}{6} + C$.

$y = \dfrac{x^3}{6} + \dfrac{C}{x^3}$.

$y(1) = 0$: $\dfrac{1}{6} + C = 0 \implies C = -\dfrac{1}{6}$.

$\boxed{y = \dfrac{x^3}{6} - \dfrac{1}{6x^3} = \dfrac{x^6 - 1}{6x^3}}$.

If you get this wrong, revise: Integrating Factor — Section 2.

Problem 3

Solve $y'' - 6y' + 9y = 0$ given $y(0) = 1$ and $y'(0) = 0$.

Solution 3

Auxiliary: $m^2 - 6m + 9 = 0 \implies (m-3)^2 = 0 \implies m = 3$ (repeated).

$y = (A + Bx)e^{3x}$, $y' = (B + 3A + 3Bx)e^{3x}$.

$y(0) = 1$: $A = 1$. $y'(0) = 0$: $B + 3 = 0 \implies B = -3$.

$\boxed{y = (1 - 3x)e^{3x}}$.

If you get this wrong, revise: Three Cases — Section 3.3.

Problem 4

Solve $y'' + 4y' + 13y = 0$.

Solution 4

Auxiliary: $m^2 + 4m + 13 = 0 \implies m = \dfrac◆LB◆-4 \pm \sqrt{16-52}◆RB◆◆LB◆2◆RB◆ = -2 \pm 3i$.

$\alpha = -2$, $\beta = 3$.

$\boxed{y = e^{-2x}(A\cos 3x + B\sin 3x)}$.

If you get this wrong, revise: Three Cases — Section 3.3.

Problem 5

Find the general solution of $y'' - y = 2e^x$.

Solution 5

CF: $m^2 - 1 = 0 \implies m = \pm 1$. $y_h = Ae^x + Be^{-x}$.

PI: $f(x) = 2e^x$. Since $e^x$ appears in the CF, try $y_p = cxe^x$.

$y_p' = c(1+x)e^x$, $y_p'' = c(2+x)e^x$.

$y_p'' - y_p = c(2+x)e^x - cxe^x = 2ce^x = 2e^x \implies c = 1$.

$\boxed{y = Ae^x + Be^{-x} + xe^x}$.

If you get this wrong, revise: Undetermined Coefficients — Section 4.2.

Problem 6

Find the general solution of $y'' + 2y' + y = x^2$.

Solution 6

CF: $m^2 + 2m + 1 = 0 \implies (m+1)^2 = 0$. $y_h = (A + Bx)e^{-x}$.

PI: Try $y_p = px^2 + qx + r$. $y_p' = 2px + q$, $y_p'' = 2p$.

$(2p) + 2(2px + q) + (px^2 + qx + r) = x^2$.

$px^2 + (4p+q)x + (2p+2q+r) = x^2$.

$p = 1$, $4 + q = 0 \implies q = -4$, $2 - 8 + r = 0 \implies r = 6$.

$\boxed{y = (A + Bx)e^{-x} + x^2 - 4x + 6}$.

If you get this wrong, revise: Undetermined Coefficients — Section 4.2.

Problem 7

A body cools from $80°C$ to $60°C$ in 10 minutes in surroundings at $20°C$. How long does it take to cool to $40°C$?

Solution 7

Newton's law: $T = 20 + 60e^{-kt}$ (since $T_0 = 80$, $T_a = 20$).

$T(10) = 60$: $60 = 20 + 60e^{-10k} \implies e^{-10k} = \dfrac{2}{3}$.

$T(t) = 40$: $40 = 20 + 60e^{-kt} \implies e^{-kt} = \dfrac{1}{3}$.

$\left(\dfrac{2}{3}\right)^{t/10} = \dfrac{1}{3} \implies \dfrac{t}{10}\ln\dfrac{2}{3} = \ln\dfrac{1}{3}$.

$t = \dfrac◆LB◆10\ln(1/3)◆RB◆◆LB◆\ln(2/3)◆RB◆ = \dfrac◆LB◆10\ln 3◆RB◆◆LB◆\ln(3/2)◆RB◆ \approx \dfrac◆LB◆10 \times 1.0986◆RB◆◆LB◆0.4055◆RB◆ \approx 27.1$ minutes.

If you get this wrong, revise: Newton's Law of Cooling — Section 5.2.

Problem 8

Solve $\dfrac{dy}{dx} = xy$ given $y(0) = 5$.

Solution 8

Separate: $\dfrac{1}{y}\,dy = x\,dx$.

Integrate: $\ln|y| = \dfrac{x^2}{2} + C$.

$y(0) = 5$: $\ln 5 = C$.

$y = e^{x^2/2 + \ln 5} = 5e^{x^2/2}$.

$\boxed{y = 5e^{x^2/2}}$.

If you get this wrong, revise: Separable Equations — Section 1.

Problem 9

Solve $y'' + 9y = 6\cos 3x$.

Solution 9

CF: $m^2 + 9 = 0 \implies m = \pm 3i$. $y_h = A\cos 3x + B\sin 3x$.

PI: $f(x) = 6\cos 3x$. Since $\cos 3x$ and $\sin 3x$ are in the CF, try $y_p = x(c\cos 3x + d\sin 3x)$.

$y_p' = (c\cos 3x + d\sin 3x) + x(-3c\sin 3x + 3d\cos 3x)$ $y_p'' = (-6c\sin 3x + 6d\cos 3x) + x(-9c\cos 3x - 9d\sin 3x)$

$y_p'' + 9y_p = -6c\sin 3x + 6d\cos 3x = 6\cos 3x$.

$d = 1$, $c = 0$.

$\boxed{y = A\cos 3x + B\sin 3x + x\sin 3x}$.

If you get this wrong, revise: Undetermined Coefficients — Section 4.2.

Problem 10

Solve $\dfrac{dy}{dx} - \dfrac{y}{x} = x^2$ for $x > 0$, given $y(1) = 3$.

Solution 10

$P(x) = -1/x$, $Q(x) = x^2$.

$\mu = e^{\int -1/x\,dx} = e^{-\ln x} = \dfrac{1}{x}$.

$\dfrac{d}{dx}\!\left(\dfrac{y}{x}\right) = \dfrac{1}{x} \cdot x^2 = x$.

$\dfrac{y}{x} = \dfrac{x^2}{2} + C \implies y = \dfrac{x^3}{2} + Cx$.

$y(1) = 3$: $\dfrac{1}{2} + C = 3 \implies C = \dfrac{5}{2}$.

$\boxed{y = \dfrac{x^3}{2} + \dfrac{5x}{2} = \dfrac{x(x^2 + 5)}{2}}$.

If you get this wrong, revise: Integrating Factor — Section 2.

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8. Advanced Worked Examples

Example 8.1: Second-order linear ODE with complex roots

Problem. Solve $\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} + 13y = 0$ given $y(0) = 2$ and $y'(0) = -1$.

Solution. Auxiliary equation: $m^2 + 4m + 13 = 0$.

$m = \dfrac◆LB◆-4 \pm \sqrt{16-52}◆RB◆◆LB◆2◆RB◆ = -2 \pm 3i$.

General solution: $y = e^{-2x}(A\cos 3x + B\sin 3x)$.

$y(0) = A = 2$.

$y' = e^{-2x}(-2A\cos 3x - 2B\sin 3x - 3A\sin 3x + 3B\cos 3x)$.

$y'(0) = -2A + 3B = -1 \implies -4 + 3B = -1 \implies B = 1$.

$\boxed{y = e^{-2x}(2\cos 3x + \sin 3x)}$

Example 8.2: Integrating factor with a tricky integral

Problem. Solve $\dfrac{dy}{dx} + \dfrac{2y}{x} = x^2$ for $x > 0$.

Solution. Integrating factor: $\mu = \exp\!\left(\displaystyle\int \frac{2}{x}\,dx\right) = e^{2\ln x} = x^2$.

$\frac{d}{dx}(x^2 y) = x^4$

$x^2 y = \frac{x^5}{5} + C$

$\boxed{y = \frac{x^3}{5} + \frac{C}{x^2}}$

Example 8.3: Homogeneous equation via substitution

Problem. Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$.

Solution. This is a homogeneous equation. Let $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.

$v + x\frac{dv}{dx} = \frac{1+v}{1-v}$

$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$

$\int \frac{1-v}{1+v^2}\,dv = \int \frac{1}{x}\,dx$

$\int \frac{1}{1+v^2}\,dv - \int \frac{v}{1+v^2}\,dv = \ln|x| + C$

$\arctan v - \frac{1}{2}\ln(1+v^2) = \ln|x| + C$

Substituting $v = y/x$:

$\arctan\!\left(\frac{y}{x}\right) - \frac{1}{2}\ln\!\left(1+\frac{y^2}{x^2}\right) = \ln|x| + C$

Example 8.4: Coupled first-order ODEs

Problem. Solve $\dfrac{dx}{dt} = 3x + 2y$, $\dfrac{dy}{dt} = -5x - y$.

Solution. From the second equation: $y = \dfrac{1}{2}\!\left(\dfrac{dx}{dt} - 3x\right)$.

Differentiating: $\dfrac{dy}{dt} = \dfrac{1}{2}\!\left(\dfrac{d^2x}{dt^2} - 3\dfrac{dx}{dt}\right)$.

Substituting into the second equation: $\dfrac{1}{2}\!\left(\dfrac{d^2x}{dt^2} - 3\dfrac{dx}{dt}\right) = -5x - \dfrac{1}{2}\!\left(\dfrac{dx}{dt} - 3x\right)$.

$\frac{d^2x}{dt^2} - 3\frac{dx}{dt} = -10x - \frac{dx}{dt} + 3x$

$\frac{d^2x}{dt^2} - 2\frac{dx}{dt} + 7x = 0$

Auxiliary: $m^2 - 2m + 7 = 0 \implies m = 1 \pm i\sqrt{6}$.

$x = e^t(A\cos\sqrt{6}\,t + B\sin\sqrt{6}\,t)$.

Then $y = \dfrac{1}{2}\!\left(\dfrac{dx}{dt} - 3x\right)$.

Example 8.5: Exponential growth with harvesting

Problem. A population $P(t)$ satisfies $\dfrac{dP}{dt} = 0.1P - 50$. Find the general solution and interpret.

Solution. This is a first-order linear ODE: $\dfrac{dP}{dt} - 0.1P = -50$.

Integrating factor: $\mu = e^{-0.1t}$.

$\frac{d}{dt}(Pe^{-0.1t}) = -50e^{-0.1t}$

$Pe^{-0.1t} = 500e^{-0.1t} + C$

$\boxed{P = 500 + Ce^{0.1t}}$

The equilibrium population is $P = 500$. If $P(0) > 500$, the population grows exponentially; if $P(0) < 500$, it decays to zero (extinction).

Example 8.6: Boundary value problem

Problem. Solve $y'' - 6y' + 9y = 0$ with $y(0) = 1$ and $y(1) = e^3$.

Solution. Auxiliary: $m^2 - 6m + 9 = 0 \implies (m-3)^2 = 0$. Repeated root $m = 3$.

General solution: $y = (A + Bx)e^{3x}$.

$y(0) = A = 1$.

$y(1) = (1+B)e^3 = e^3 \implies 1+B = 1 \implies B = 0$.

$\boxed{y = e^{3x}}$

Example 8.7: Non-homogeneous second-order ODE

Problem. Solve $y'' + y = 2\cos x$.

Solution. Complementary function: $m^2 + 1 = 0 \implies m = \pm i$. $y_c = A\cos x + B\sin x$.

Particular integral: Since $\cos x$ is part of the CF, try $y_p = x(C\cos x + D\sin x)$.

$y_p' = C\cos x + D\sin x + x(-C\sin x + D\cos x)$.

$y_p'' = -C\sin x + D\cos x + (-C\sin x + D\cos x) + x(-C\cos x - D\sin x) = 2(-C\sin x + D\cos x) - x(C\cos x + D\sin x)$.

$y_p'' + y_p = 2(-C\sin x + D\cos x) = 2\cos x$.

$D = 1$, $C = 0$. So $y_p = x\sin x$.

$\boxed{y = A\cos x + B\sin x + x\sin x}$

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9. Common Pitfalls

Pitfall	Correct Approach
Forgetting the constant of integration when using an integrating factor	Integrate both sides after multiplying by $\mu$; $+C$ appears on the right
Using the wrong trial function for the particular integral	If the RHS is part of the complementary function, multiply by $x$ (or $x^2$ if needed)
Confusing $Ae^{mx}$ (single root) with $(A+Bx)e^{mx}$ (repeated root)	Check the discriminant: repeated root $\iff$ discriminant $= 0$
Not applying initial conditions to find $A$ and $B$	Always substitute the given conditions into the general solution and its derivative

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10. Additional Exam-Style Questions

Question 8

Solve $\dfrac{d^2y}{dx^2} - 2\dfrac{dy}{dx} - 3y = 6e^{2x}$.

Solution

CF: $m^2 - 2m - 3 = 0 \implies (m-3)(m+1) = 0$. $y_c = Ae^{3x} + Be^{-x}$.

PI: Try $y_p = Ce^{2x}$. Substituting: $4C - 4C - 3C = 6 \implies C = -2$.

$\boxed{y = Ae^{3x} + Be^{-x} - 2e^{2x}}$

Question 9

A body cools according to $\dfrac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ = -k(\theta - 20)$, where $\theta$ is the temperature in $°\mathrm{C}$ and $20°\mathrm{C}$ is the room temperature. If $\theta(0) = 90$ and $\theta(10) = 50$, find $\theta(30)$.

Solution

$\dfrac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ + k\theta = 20k$. IF: $e^{kt}$.

$\theta e^{kt} = 20e^{kt} + C$

$\theta = 20 + Ce^{-kt}$.

$\theta(0) = 90 \implies C = 70$. $\theta(10) = 50 \implies 50 = 20 + 70e^{-10k} \implies e^{-10k} = \dfrac{3}{7}$.

$\theta(30) = 20 + 70e^{-30k} = 20 + 70\!\left(\dfrac{3}{7}\right)^{\!3} = 20 + 70 \times \dfrac{27}{343} = 20 + \dfrac{1890}{343}$.

$\boxed{\theta(30) \approx 25.5°\mathrm{C}}$

Question 10

Prove that the substitution $y = vx$ transforms $\dfrac{dy}{dx} = f\!\left(\dfrac{y}{x}\right)$ into a separable equation.

Solution

Let $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.

Substituting: $v + x\dfrac{dv}{dx} = f(v)$.

$x\frac{dv}{dx} = f(v) - v$

$\frac{dv}{f(v) - v} = \frac{dx}{x}$

This is separable. Integrating gives $\displaystyle\int \frac{dv}{f(v) - v} = \ln|x| + C$. $\blacksquare$

Question 11

Find the particular solution to $x\dfrac{dy}{dx} - y = x^2$ with $y(1) = 3$.

Solution

$\dfrac{dy}{dx} - \dfrac{y}{x} = x$. IF: $\mu = \exp\!\left(\displaystyle\int -\dfrac{1}{x}\,dx\right) = \dfrac{1}{x}$.

$\frac{1}{x}\cdot\frac{dy}{dx} - \frac{y}{x^2} = 1$

$\frac{d}{dx}\!\left(\frac{y}{x}\right) = 1$

$\frac{y}{x} = x + C \implies y = x^2 + Cx$

$y(1) = 1 + C = 3 \implies C = 2$.

$\boxed{y = x^2 + 2x}$

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11. Connections to Other Topics

11.1 Differential equations and calculus

Solving ODEs requires integration techniques (substitution, parts, partial fractions). See Further Calculus.

11.2 Second-order ODEs and complex numbers

The auxiliary equation uses complex roots to give oscillatory solutions $e^{\alpha t}(A\cos\beta t + B\sin\beta t)$. See Complex Numbers.

11.3 Differential equations and mechanics

Newton's second law $F = ma$ leads to second-order ODEs in mechanics. See Circular Motion.

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12. Key Results Summary

ODE Type	Method	General Solution
$\dfrac{dy}{dx} + P(x)y = Q(x)$	Integrating factor $\mu = e^{\int P\,dx}$	$y = \dfrac◆LB◆1◆RB◆◆LB◆\mu◆RB◆\displaystyle\int \mu Q\,dx$
$\dfrac{dy}{dx} = f\!\left(\dfrac{y}{x}\right)$	Substitution $y = vx$	Separate and integrate
$a\dfrac{d^2y}{dx^2}+b\dfrac{dy}{dx}+cy=0$	Auxiliary equation $am^2+bm+c=0$	Real roots: $Ae^{m_1x}+Be^{m_2x}$; repeated: $(A+Bx)e^{mx}$; complex: $e^{\alpha x}(A\cos\beta x+B\sin\beta x)$

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13. Further Exam-Style Questions

Question 12

Solve $\dfrac{d^2y}{dx^2} - 6\dfrac{dy}{dx} + 25y = 0$ with $y(0) = 0$ and $y'(0) = 3$.

Solution

$m^2 - 6m + 25 = 0 \implies m = 3 \pm 4i$.

$y = e^{3x}(A\cos 4x + B\sin 4x)$.

$y(0) = A = 0$. $y' = 3e^{3x}B\sin 4x + 4e^{3x}B\cos 4x$. $y'(0) = 4B = 3 \implies B = 3/4$.

$\boxed{y = \dfrac{3}{4}e^{3x}\sin 4x}$

Question 13

Prove that the Wronskian $W(y_1, y_2) = y_1y_2' - y_1'y_2 \neq 0$ if and only if $y_1$ and $y_2$ are linearly independent solutions of a second-order linear ODE.

Solution

If $y_1$ and $y_2$ are linearly dependent, $y_2 = cy_1$, then $W = y_1(cy_1') - y_1'(cy_1) = 0$.

Conversely, if $W = 0$ at some point and both satisfy the same linear ODE, then the initial value problem with initial conditions matching $y_1$ and $y_2$ would have two solutions, contradicting uniqueness. Hence $y_1$ and $y_2$ must be linearly dependent.

Therefore $W \neq 0 \iff$ linearly independent. $\blacksquare$

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14. Advanced Topics

14.1 The integrating factor method — derivation

For $\dfrac{dy}{dx} + P(x)y = Q(x)$, multiply by $\mu = e^{\int P\,dx}$:

$\mu\frac{dy}{dx} + \mu Py = \mu Q$

$\frac{d}{dx}(\mu y) = \mu Q$

$\mu y = \int \mu Q\,dx + C$

$y = \frac◆LB◆1◆RB◆◆LB◆\mu◆RB◆\int \mu Q\,dx + \frac◆LB◆C◆RB◆◆LB◆\mu◆RB◆$

14.2 Systems of linear ODEs

For the system $\dot{\mathbf{x}} = \mathbf{A}\mathbf{x}$ where $\mathbf{x} = (x_1, \ldots, x_n)$:

If $\mathbf{A}$ is diagonalisable with $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, let $\mathbf{z} = \mathbf{P}^{-1}\mathbf{x}$:

$\dot{\mathbf{z}} = \mathbf{D}\mathbf{z}$, giving $z_i = c_i e^{\lambda_i t}$.

$\mathbf{x} = \mathbf{P}\mathbf{z} = \sum c_i e^{\lambda_i t}\mathbf{v}_i$.

14.3 Boundary value problems vs initial value problems

An IVP specifies $y$ and $y'$ at one point. A BVP specifies $y$ at two (or more) points. BVPs may have zero, one, or multiple solutions, unlike IVPs which (for linear ODEs) have a unique solution.

14.4 Phase portraits

For autonomous 2D systems $\dot{x} = f(x,y)$, $\dot{y} = g(x,y)$, the phase portrait shows trajectories in the $xy$-plane. Key features:

• Fixed points: where $\dot{x} = \dot{y} = 0$
• Stability: determined by the eigenvalues of the Jacobian at each fixed point

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15. Further Exam-Style Questions

Question 14

Solve $\dfrac{dy}{dx} = \dfrac{x^2+y^2}{xy}$ using an appropriate substitution.

Solution

This is homogeneous: $\dfrac{dy}{dx} = \dfrac{1+(y/x)^2}{y/x}$. Let $v = y/x$:

$v + x\dfrac{dv}{dx} = \dfrac{1+v^2}{v} = \dfrac{1}{v} + v$.

$x\dfrac{dv}{dx} = \dfrac{1}{v}$.

$\int v\,dv = \int \dfrac{dx}{x} \implies \dfrac{v^2}{2} = \ln|x| + C$.

$\dfrac{y^2}{2x^2} = \ln|x| + C$.

$y^2 = 2x^2(\ln|x|+C)$.

Question 15

Prove that the general solution of $\dfrac{d^2y}{dx^2} + \omega^2 y = 0$ can be written as $y = R\cos(\omega x - \delta)$ where $R$ and $\delta$ are constants.

Solution

The general solution is $y = A\cos\omega x + B\sin\omega x$.

Let $R\cos\delta = A$ and $R\sin\delta = B$. Then $R = \sqrt{A^2+B^2}$ and $\delta = \arctan(B/A)$.

$A\cos\omega x + B\sin\omega x = R\cos\delta\cos\omega x + R\sin\delta\sin\omega x = R\cos(\omega x - \delta)$. $\blacksquare$