Differential Equations

Differential Equations

Differential equations are equations involving derivatives of an unknown function. They arise 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.

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

Board Coverage

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

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 $A \to \infty$.

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:

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

Since $\dfrac{d\mu}{dx} = \mu \cdot P(x)$ (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:

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

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

$\square$

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

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

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

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

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.

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 $y_p = px + q$. $y_p' = p$, $y_p'' = 0$.

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

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

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

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 $\dfrac{d\theta}{dt} = -k\theta$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.

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.

$\square$

Problems

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: $m^2 + 4m + 13 = 0$.

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

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