Study Notes on Linear and Bernoulli Equations and Second-Order ODEs

First-Order Linear Ordinary Differential Equations (ODEs)

Definition of a First-Order Linear ODE: A first-order linear ordinary differential equation is defined by its standard mathematical form:
- $\frac{dy}{dx} + P(x) y = Q(x)$
- In this context, $P(x)$ and $Q(x)$ are given continuous functions of the independent variable $x$ .
Integrating Factor Method (Theorem 4.1): This method identifies a specific function, known as the integrating factor (IF), denoted as $\mu(x)$ , which simplifies the ODE into a solvable perfect derivative.
- The formula for the integrating factor is: $\mu(x) = e^{\int P(x) \,dx}$ .
- When the standard form equation is multiplied by $\mu(x)$ , the left-hand side becomes: $\frac{d}{dx} [\mu(x) y] = \mu(x) Q(x)$ .
- By integrating both sides with respect to $x$ , the general solution is obtained: $y \cdot e^{\int P(x) \,dx} = \int Q(x) e^{\int P(x) \,dx} \,dx + C$ .
Step-by-Step Procedure for Solving First-Order Linear ODEs:
1. Standardize: Write the given ODE in the standard form: $\frac{dy}{dx} + P(x)y = Q(x)$ .
2. Calculate Integrated Exponent: Compute the integrating factor: $\mu = e^{\int P(x) \,dx}$ .
3. Distribute: Multiply the entire equation by the calculated $\mu$ .
4. Identify Derivative: Recognize that the left side of the equation is now equal to $\frac{d(\mu y)}{dx}$ .
5. Integrate: Integrate both sides of the equation: $\mu y = \int \mu Q(x) \,dx + C$ .
6. Isolate Variable: Divide by $\mu$ to solve for the dependent variable $y$ .
Example 4.1: Solve $x \frac{dy}{dx} + y = x^3$
- Divide through by $x$ to standardize: $\frac{dy}{dx} + (\frac{1}{x})y = x^2$ .
- Identify components: $P(x) = \frac{1}{x}$ and $Q(x) = x^2$ .
- Calculate IF: $\mu = e^{\int (\frac{1}{x}) \,dx} = e^{\ln(x)} = x$ .
- Multiply through by $x$ : $\frac{d(xy)}{dx} = x^3$ .
- Integrate both sides: $xy = \frac{x^4}{4} + C$ .
- Final Solution: $y = \frac{x^3}{4} + \frac{C}{x}$ .
Example 4.2: Solve $\frac{dy}{dx} - 2y = 4$
- Identify components: $P(x) = -2$ and $Q(x) = 4$ .
- Calculate IF: $\mu = e^{\int (-2) \,dx} = e^{-2x}$ .
- Multiply through: $\frac{d(y e^{-2x})}{dx} = 4 e^{-2x}$ .
- Integrate: $y e^{-2x} = -2 e^{-2x} + C$ .
- General Solution: $y = -2 + C e^{2x}$ .
Example 4.3: Solve $x^2 \frac{dy}{dx} - 3y = \sqrt{x}$
- Standardize by dividing by $x^2$ : $\frac{dy}{dx} - (\frac{3}{x^2})y = x^{-3/2}$ .
- Identify components: $P(x) = -\frac{3}{x^2}$ .
- Calculate IF: $\mu = e^{\int -3x^{-2} \,dx} = e^{3/x}$ .
- Multiply through: $\frac{d(y e^{3/x})}{dx} = x^{-3/2} e^{3/x}$ .
- The solution is then obtained by integrating the right-hand side using appropriate methods such as substitution or integration by parts.

Bernoulli Differential Equations

Definition of a Bernoulli Equation: A Bernoulli equation is a non-linear first-order ODE expressed as:
- $\frac{dy}{dx} + P(x) y = Q(x) y^n$
- The variable $n$ is a real number.
- If $n = 0$ , the equation is linear.
- If $n = 1$ , the equation is separable.
- For all other values where $n \neq 0, 1$ , the equation is non-linear but can be linearized via substitution.
Reduction to Linear Form (Theorem 4.2):
- Divide the entire equation by $y^n$ : $y^{-n} \frac{dy}{dx} + P(x) y^{1-n} = Q(x)$ .
- Introduce a new variable: Let $z = y^{1-n}$ .
- Determine the derivative of the substitution: $\frac{dz}{dx} = (1-n) y^{-n} \frac{dy}{dx}$ .
- Substitute these into the equation to transform it into a linear ODE: $\frac{dz}{dx} + (1-n) P(x) z = (1-n) Q(x)$ .
- Solve for $z$ , then recover $y$ using the relation $z = y^{1-n}$ .
Step-by-Step Procedure for Bernoulli Equations:
1. Identify the values of $P(x)$ , $Q(x)$ , and the exponent $n$ from the initial ODE.
2. Divide the equation by $y^n$ to reach the form: $y^{-n} \frac{dy}{dx} + P(x) y^{1-n} = Q(x)$ .
3. Apply the substitution $z = y^{1-n}$ and compute $\frac{dz}{dx} = (1-n) y^{-n} \frac{dy}{dx}$ .
4. Multiply the equation by $\text{(1 – n)}$ to establish the linear form: $\frac{dz}{dx} + (1-n) P(x) z = (1-n) Q(x)$ .
5. Solve this linear ODE utilizing the Integrating Factor method.
6. Back-substitute to find the final solution for $y$ .
Example 4.4: Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{y^3}{x}$
- Identify parameters: $n = 3$ , $P(x) = \frac{1}{x}$ , and $Q(x) = \frac{1}{x}$ .
- Divide by $y^3$ : $y^{-3} \frac{dy}{dx} + (\frac{1}{x}) y^{-2} = \frac{1}{x}$ .
- Substitute $z = y^{-2}$ , meaning $\frac{dz}{dx} = -2 y^{-3} \frac{dy}{dx}$ .
- Multiply by $-2$ to linearize: $\frac{dz}{dx} - (\frac{2}{x})z = -\frac{2}{x}$ .
- Solve linearized ODE ( $P = -\frac{2}{x}$ , $Q = -\frac{2}{x}$ ):
  - $\mu = e^{\int -\frac{2}{x} \,dx} = x^{-2}$ .
  - $\frac{d(x^{-2} z)}{dx} = -\frac{2}{x^3}$ .
  - $x^{-2} z = \frac{1}{x^2} + C$ .
  - $z = 1 + Cx^2$ .
- Recover $y$ : Since $z = y^{-2}$ , then $y^{-2} = 1 + Cx^2$ , leading to $y = \frac{1}{\sqrt{1 + Cx^2}}$ .
Example 4.5: Solve $\frac{dy}{dx} - 2xy = 2xy^3$
- Identify parameters: $n = 3$ , $P(x) = -2x$ , $Q(x) = 2x$ .
- Let $z = y^{-2}$ , hence $\frac{dz}{dx} = -2y^{-3}\frac{dy}{dx}$ .
- Multiply by $-2y^{-3}$ : $\frac{dz}{dx} + 4xz = -4x$ .
- Calculate IF: $\mu = e^{\int 4x \,dx} = e^{2x^2}$ .
- Set up derivative: $\frac{d(z e^{2x^2})}{dx} = -4x e^{2x^2}$ .
- Integrate: $z e^{2x^2} = -4 \int x e^{2x^2} \,dx = -e^{2x^2} + C$ .
- Solve for $z$ : $z = -1 + C e^{-2x^2}$ .
- Final Solution: $y^{-2} = -1 + C e^{-2x^2}$ or $1 = y^2 (C e^{-2x^2} - 1)$ .

Tutor-Marked Assignment — Section 2, Unit 2

Part 1: Solve the following linear ODEs:
- (i) $x \frac{dy}{dx} + y = x^3$
- (ii) $x \frac{dy}{dx} - 5y = x^2$
- (iii) $x^2 \frac{dy}{dx} - 3y = \sqrt{x}$
- (iv) $\frac{dy}{dx} + y \cot(x) = \cos(x)$
- (v) (1 - x^2) \frac{dy}{dx} - xy = 1
Part 2: Solve the following Bernoulli equations:
- (i) $x^2 y - x^3 \frac{dy}{dx} = 4y^4 \cos(x)$
- (ii) $2y - 3 \frac{dy}{dx} = y^4 e^{3x}$
- (iii) $y - 2x \frac{dy}{dx} = x(x+1)y^3$
- (iv) $\frac{dy}{dx} + y = xy^3$
- (v) $\frac{dy}{dx} - y = x$
- (vi) $x \frac{dy}{dx} + 2y = \frac{1}{x^3}$
- (vii) $\frac{dy}{dx} - 2xy = 6y^3 x^2$

Solution of Second-Order Ordinary Differential Equations

Context and Application: Second-order ODEs are fundamental in modeling physical systems including mechanics (e.g., spring-mass systems) and electrical circuits. This study focuses specifically on second-order linear ODEs with constant coefficients.
General Form of Second-Order Linear ODEs (Definition 5.1):
- A general second-order linear ODE is written as: $a(x) y'' + b(x) y' + c(x) y = f(x)$ .
- The terms $a(x)$ , $b(x)$ , $c(x)$ , and $f(x)$ , are known functions of $x$ .
- Homogeneous vs. Non-Homogeneous:
  - If $f(x) \equiv 0$ , the equation is classified as homogeneous.
  - If $f(x) \neq 0$ , the equation is non-homogeneous.
- Constant Coefficients: If the functions $a, b, c$ are constants rather than functions of $x$ , the equation is simplified to: $a y'' + b y' + c y = f(x)$ .
- The homogeneous part of this constant coefficient equation is: $a y'' + b y' + c y = 0$ .
The Auxiliary (Characteristic) Equation:
- To solve the constant-coefficient homogeneous equation, a trial solution of the form $y = e^{mx}$ is used.
- Differentiating yields $y' = m e^{mx}$ and $y'' = m^2 e^{mx}$ .
- Substituting these into the equation results in: $a(m^2 e^{mx}) + b(m e^{mx}) + c(e^{mx}) = 0$ .
- Since $e^{mx}$ is never zero, we divide by it to obtain the Auxiliary Equation: $a m^2 + b m + c = 0$ .
- The roots of this quadratic equation ( $m_1$ and $m_2$ ) define the form of the Complementary Function (C.F.), denoted as $y_c$ .

Cases of the Auxiliary Equation

Case I: Two Distinct Real Roots ( $m_1 \neq m_2$ , both real):
- If the auxiliary equation provides two unique real roots, the solution is the sum of two exponential terms.
- General Solution: $y_c = c_1 e^{m_1 x} + c_2 e^{m_2 x}$ , where $c_1$ and $c_2$ are arbitrary constants.
- Example 5.1: Solve $y'' + 5y' + 6y = 0$
  - Auxiliary equation: $m^2 + 5m + 6 = 0$ .
  - Factoring: $(m + 2)(m + 3) = 0 \implies m_1 = -2, m_2 = -3$ .
  - Solution: $y = c_1 e^{-2x} + c_2 e^{-3x}$ .
- Example 5.2: Solve $y'' - 3y' - 4y = 0$
  - Auxiliary equation: $m^2 - 3m - 4 = 0$ .
  - Factoring: $(m - 4)(m + 1) = 0 \implies m_1 = 4, m_2 = -1$ .
  - Solution: $y = c_1 e^{4x} + c_2 e^{-x}$ .
Case II: Repeated (Equal) Real Roots ( $m_1 = m_2 = m$ ):
- If there is a repeated real root of multiplicity 2, an additional factor of $x$ is introduced to ensure the solutions remain linearly independent.
- General Solution: $y_c = (c_1 + c_2 x) e^{mx}$ .
- Example 5.3: Solve $y'' - 6y' + 9y = 0$
  - Auxiliary equation: $m^2 - 6m + 9 = 0$ .
  - Factoring: $(m - 3)^2 = 0 \implies m = 3$ .
  - Solution: $y = (c_1 + c_2 x) e^{3x}$ .
- Example 5.4: Solve $y'' + 4y' + 4y = 0$
  - Auxiliary equation: $m^2 + 4m + 4 = 0$ .
  - Factoring: $(m + 2)^2 = 0 \implies m = -2$ .
  - Solution: $y = (c_1 + c_2 x) e^{-2x}$ .
Case III: Complex Conjugate Roots ( $m = \alpha \pm i\beta$ ):
- If the roots are complex conjugate pairs, where $\alpha$ is the real part and $\beta$ is the imaginary part ( $\beta \neq 0$ ), the solution utilizes Euler's formula: $e^{i\beta x} = \cos(\beta x) + i \sin(\beta x)$ .
- General Solution: $y_c = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))$ .
- Special Case - Purely Imaginary Roots: If $\alpha = 0$ , the solution is simply: $y_c = c_1 \cos(\beta x) + c_2 \sin(\beta x)$ .
- Example 5.5: Solve $y'' + 4y' + 13y = 0$
  - Auxiliary equation: $m^2 + 4m + 13 = 0$ .
  - Quadratic formula: $m = \frac{-4 \pm \sqrt{16-52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i$ .
  - Identity: $\alpha = -2, \beta = 3$ .
  - General solution: $y = e^{-2x} (c_1 \cos(3x) + c_2 \sin(3x))$ .
- Example 5.6: Solve $y'' + 4y = 0$
  - Auxiliary equation: $m^2 + 4 = 0 \implies m = \pm 2i$ .
  - Identity: $\alpha = 0, \beta = 2$ .
  - General solution: $y = c_1 \cos(2x) + c_2 \sin(2x)$ .