Differential Equations

DIFFERENTIAL EQUATIONS — COMPREHENSIVE CHAPTER NOTES

Introduction and Definition

A **differential equation** is an equation involving derivatives of a dependent variable with respect to one or more independent variables. It arises naturally in modelling real-world phenomena in physics, chemistry, biology, economics, and engineering.

**Key distinction**: Unlike algebraic equations where solutions are numbers, solutions to differential equations are functions.

**Example**: The equation dy/dx = x + y involves the derivative dy/dx, making it a differential equation.

An **ordinary differential equation (ODE)** involves derivatives with respect to only one independent variable. A **partial differential equation (PDE)** involves derivatives with respect to multiple independent variables. In this chapter, we study only ordinary differential equations.

---

Order of a Differential Equation

The **order** of a differential equation is the order of the highest derivative present in the equation.

**Rules for identifying order:**

Find the highest derivative in the equation

The power of that derivative gives the order

Order is always a positive integer

**Examples:**

dy/dx = e^x has order 1 (first-order derivative dy/dx is highest)

d²y/dx² + y = 0 has order 2 (second-order derivative d²y/dx² is highest)

d³y/dx³ + 2d²y/dx² + dy/dx = 0 has order 3

**Notation**: For nth order derivatives, use the symbol y^(n) or d^n y/dx^n.

---

Degree of a Differential Equation

The **degree** of a differential equation is the highest power (positive integral exponent) of the highest-order derivative, **provided the equation is a polynomial equation in all its derivatives**.

**Important condition**: The equation must be polynomial in its derivatives (not involving trigonometric, exponential, or other non-polynomial functions of derivatives).

**When degree is NOT defined:**

If the equation is not polynomial in derivatives

If any derivative appears inside a trigonometric, exponential, or non-polynomial function

Example: sin(dy/dx) + y = 0 has no defined degree

**Examples:**

(d²y/dx²)³ + (dy/dx)² + y = 0 has order 2, degree 3 (highest power of d²y/dx² is 3)

d³y/dx³ + 2d²y/dx² + dy/dx = 5 has order 3, degree 1

cos(d²y/dx²) + y = 0 has order 2, degree not defined (polynomial condition violated)

**Key fact**: Order and degree (when defined) are always positive integers.

---

General and Particular Solutions

**General solution (primitive)**: A solution containing arbitrary constants. The number of arbitrary constants equals the order of the differential equation.

A first-order DE has one arbitrary constant

A second-order DE has two arbitrary constants

An nth-order DE has n arbitrary constants

**Particular solution**: A solution obtained from the general solution by assigning specific values to the arbitrary constants, usually using given initial or boundary conditions.

**Verification**: To verify a function is a solution:

1. Differentiate the function as many times as the order of the DE

2. Substitute into the original equation

3. Check if LHS = RHS

**Example: Verify y = a cos x + b sin x solves d²y/dx² + y = 0**

y = a cos x + b sin x

dy/dx = −a sin x + b cos x

d²y/dx² = −a cos x − b sin x = −y

Substituting: d²y/dx² + y = −y + y = 0 ✓

This is the general solution (contains two constants a and b for a second-order equation).

If we set a = 1, b = 0, we get y = cos x, a particular solution.

---

Formation of Differential Equations

To form a DE from a given family of curves:

1. Write the equation of the family: F(x, y, c₁, c₂, ..., cₙ) = 0

2. Differentiate successively n times (where n = number of constants)

3. Eliminate the arbitrary constants using the original equation and derivatives

4. The resulting equation is the differential equation

**Example: Form a DE for the family y = ae^(2x) + be^(−3x)**

y = ae^(2x) + be^(−3x) ... (1)

dy/dx = 2ae^(2x) − 3be^(−3x) ... (2)

d²y/dx² = 4ae^(2x) + 9be^(−3x) ... (3)

From (1): ae^(2x) + be^(−3x) = y

From (3): 4ae^(2x) + 9be^(−3x) = d²y/dx²

Multiply (1) by 4: 4ae^(2x) + 4be^(−3x) = 4y

Subtract: 5be^(−3x) = d²y/dx² − 4y

From (2): 2ae^(2x) − 3be^(−3x) = dy/dx

Solving these gives: **d²y/dx² − dy/dx − 6y = 0**

---

Variables Separable Method

This is the simplest method for solving first-order, first-degree DEs of the form dy/dx = F(x, y).

**Condition for separability**: F(x, y) can be written as a product g(x)·h(y), where g depends only on x and h depends only on y.

**Solution process:**

1. Write the DE as: dy/dx = g(x)·h(y)

2. If h(y) ≠ 0, separate: [1/h(y)]dy = g(x)dx

3. Integrate both sides: ∫[1/h(y)]dy = ∫g(x)dx + C

4. Result is: H(y) = G(x) + C, where H and G are antiderivatives

**Example 1: Solve dy/dx = (1 + x)/(2 − y), y ≠ 2**

Separate: (2 − y)dy = (1 + x)dx

Integrate: ∫(2 − y)dy = ∫(1 + x)dx

2y − y²/2 = x + x²/2 + C

General solution: **x² + y² + 2x − 4y + C = 0**

**Example 2: Solve dy/dx = (1 + y²)/(1 + x²)**

Separate: dy/(1 + y²) = dx/(1 + x²)

Integrate: ∫dy/(1 + y²) = ∫dx/(1 + x²)

tan⁻¹y = tan⁻¹x + C

General solution: **tan⁻¹y − tan⁻¹x = C** or **y = tan(tan⁻¹x + C)**

**Example 3: Particular solution with initial condition**

Solve dy/dx = −4xy² with y = 1 when x = 0

Separate: dy/y² = −4x dx

Integrate: −1/y = −2x² + C

When x = 0, y = 1: −1 = 0 + C, so C = −1

Particular solution: **y = 1/(2x² + 1)**

**Real-life application (Exponential growth):**

If dP/dt = rP (population growth), separate: dP/P = r dt

Integrate: ln P = rt + C

P = Ae^(rt) where A = e^C

If principal increases at 5% per year (dP/dt = 0.05P), and P₀ = Rs 1000, to find when it doubles:

2000 = 1000·e^(0.05t)

2 = e^(0.05t)

t = ln(2)/0.05 ≈ 13.86 years

---

Homogeneous Differential Equations

**Homogeneous function of degree n**: A function F(x, y) such that:

F(λx, λy) = λ^n F(x, y) for any nonzero constant λ

**Examples:**

F(x, y) = x² + 2xy is homogeneous of degree 2: F(λx, λy) = λ²(x² + 2xy) = λ²F(x, y)

F(x, y) = sin(y/x) is homogeneous of degree 0: F(λx, λy) = sin(λy/λx) = sin(y/x) = λ⁰F(x, y)

F(x, y) = x + y + 1 is NOT homogeneous

**Homogeneous differential equation**: An equation dy/dx = F(x, y) where F is homogeneous of degree zero.

**Solution method using substitution y = vx:**

1. Let y = vx where v is a function of x

2. Then dy/dx = v + x(dv/dx)

3. Substitute into the original equation: v + x(dv/dx) = F(x, vx) = F(x, y)

4. Since F is homogeneous of degree 0: F(x, vx) = x⁰·F(1, v) = f(v)

5. This gives: v + x(dv/dx) = f(v)

6. Rearrange: x(dv/dx) = f(v) − v

7. Separate: dv/[f(v) − v] = dx/x

8. Integrate both sides

9. Substitute back v = y/x to get the solution in terms of x and y

**Example: Solve (x + y)dx − (x − y)dy = 0**

Rewrite: dy/dx = (x + y)/(x − y)

Check: F(λx, λy) = (λx + λy)/(λx − λy) = (x + y)/(x − y) = λ⁰F(x, y) ✓ (homogeneous of degree 0)

Let y = vx, so dy/dx = v + x(dv/dx)

v + x(dv/dx) = (x + vx)/(x − vx) = (1 + v)/(1 − v)

x(dv/dx) = (1 + v)/(1 − v) − v = (1 + v − v(1 − v))/(1 − v) = (1 + v²)/(1 − v)

Separate: (1 − v)/(1 + v²) dv = dx/x

Integrate: ∫(1 − v)/(1 + v²) dv = ∫dx/x

∫1/(1 + v²) dv − ∫v/(1 + v²) dv = ln|x| + C

tan⁻¹v − (1/2)ln(1 + v²) = ln|x| + C

Substitute v = y/x: **tan⁻¹(y/x) − (1/2)ln(1 + y²/x²) = ln|x| + C**

Or simplified: **tan⁻¹(y/x) − (1/2)ln(x² + y²) = C**

---

Linear First-Order Differential Equations

A **linear first-order differential equation** has the standard form:

**dy/dx + P(x)y = Q(x)**

where P(x) and Q(x) are functions of x alone.

**Solution using integrating factor:**

1. Write the equation in standard form: dy/dx + P(x)y = Q(x)

2. Find the **integrating factor (IF)**: IF = e^(∫P(x)dx)

3. Multiply the entire equation by IF:

e^(∫P(x)dx) · dy/dx + e^(∫P(x)dx) · P(x)y = e^(∫P(x)dx) · Q(x)

4. Observe that the left side is the derivative of [y · e^(∫P(x)dx)]:

d/dx[y · e^(∫P(x)dx)] = e^(∫P(x)dx) · Q(x)

5. Integrate both sides:

y · e^(∫P(x)dx) = ∫e^(∫P(x)dx) · Q(x) dx + C

6. Solve for y:

**y = e^(−∫P(x)dx)[∫e^(∫P(x)dx) · Q(x) dx + C]**

**Example 1: Solve dy/dx + y = e^(−x)**

Here P(x) = 1, Q(x) = e^(−x)

IF = e^(∫1 dx) = e^x

Multiply: e^x·dy/dx + e^x·y = e^x·e^(−x) = 1

d/dx[y·e^x] = 1

Integrate: y·e^x = x + C

Solution: **y = (x + C)e^(−x)**

**Example 2: Solve dy/dx − y/x = x, x ≠ 0**

Here P(x) = −1/x, Q(x) = x

IF = e^(∫−1/x dx) = e^(−ln|x|) = 1/|x| = 1/x (taking x > 0)

Multiply: (1/x)·dy/dx − (1/x²)·y = 1

d/dx[y/x] = 1

Integrate: y/x = x + C

Solution: **y = x² + Cx**

---

Applications of Differential Equations

**1. Exponential growth and decay:**

If the rate of change is proportional to the quantity present: dy/dt = ky

Solution: y = Ae^(kt) where A = y(0)

**2. Newton's Law of Cooling:**

dT/dt = −k(T − T_m) where T is temperature, T_m is surrounding temperature

Solution: T = T_m + (T₀ − T_m)e^(−kt)

**3. Population growth:**

dP/dt = rP (unrestricted)

Solution: P(t) = P₀e^(rt)

**4. Bacterial growth:**

If bacteria doubles in certain time with growth rate proportional to population, use dN/dt = kN

Example: If 100,000 bacteria increase by 10% in 2 hours, find time to reach 200,000.

N = N₀e^(kt)

110,000 = 100,000·e^(2k)

e^(2k) = 1.1

k = (ln 1.1)/2 ≈ 0.0477

For N = 200,000:

200,000 = 100,000·e^(0.0477t)

2 = e^(0.0477t)

t = ln(2)/0.0477 ≈ **14.5 hours**

**5. Radioactive decay:**

dM/dt = −λM (where λ is decay constant)

Solution: M(t) = M₀e^(−λt)

**6. Mixing problems:**

Rate of change of substance = Rate in − Rate out

---

Key Points for Board Exams

**Always identify the type** of DE before solving (variables separable, homogeneous, or linear)

**Check initial conditions** carefully for particular solutions

**Verify your solution** by substituting back into the original equation

**Simplify final answers** to the simplest form (combine logs, simplify constants)

For **homogeneous equations**, substitute y = vx systematically

For **linear equations**, the integrating factor method is always reliable

**Real-life problems** typically lead to variables-separable or exponential-growth DEs

**Degree may not be defined** — always check if the equation is polynomial in derivatives before stating degree

---

**Common mistakes to avoid:**

Confusing order and degree

Forgetting the arbitrary constant after integration

Not simplifying the integrating factor correctly

Misidentifying whether an equation is homogeneous

Arithmetic errors when combining logarithmic constants

Forgetting to substitute back v = y/x in homogeneous equations