Unit IV: Advanced Ordinary Differential Equations and Trajectories

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a systematic procedure to find a particular solution (y_p) to non-homogeneous linear ordinary differential equations with constant coefficients. This method is highly effective when the non-homogeneous term, g(x), belongs to a specific family of functions whose derivatives are finite in variety.

Applicability and Limitations

This method is applicable to differential equations of the form:

a_n dⁿy/dxⁿ + a_n-1 d^n-1y/dx^n-1 + ... + a₁ dy/dx + a₀y = g(x)

where the coefficients a_i are constants, and g(x) consists of:

• Polynomials in x (e.g., x² + 2x)
• Exponential functions (e.g., e^αx)
• Sine or cosine functions (e.g., sin(βx) or cos(βx))
• Sums or products of the above functions

Limitation: It cannot be applied if g(x) contains division by variables, logarithmic functions, tangent functions, or other non-polynomial transcendentals (e.g., ln(x), tan(x), or 1/x). For those cases, the Method of Variation of Parameters must be utilized.

Trial Particular Solutions Table

The general form of the assumed trial solution depends on the structure of g(x):

The Duplication Rule (Modification Rule)

If any term in the assumed trial solution y_p is a solution to the corresponding homogeneous differential equation (i.e., appears in the complementary function y_c), the trial solution must be multiplied by x^s, where s is the smallest positive integer that eliminates all duplication.

Step-by-Step Solution Procedure

1. Solve the Homogeneous Equation: Find the complementary function, y_c.
2. Analyze g(x): Identify the form of the non-homogeneous term and select the initial trial solution y_p.
3. Apply the Modification Rule: Check for duplication between y_c and y_p. Multiply y_p by x^s if duplication occurs.
4. Differentiate and Substitute: Find the derivatives of y_p and substitute them into the original differential equation.
5. Solve for Coefficients: Equate coefficients of like terms on both sides of the equation to find the values of the undetermined constants.
6. Form the General Solution: Combine both parts: y = y_c + y_p.

Detailed Example

Solve: d²y/dx² - 3 dy/dx + 2y = 2e^3x

Step 1: Find the complementary function (y_c).

The auxiliary equation is:

m² - 3m + 2 = 0

Factoring gives: (m - 1)(m - 2) = 0, so m = 1 or m = 2.

Thus, y_c = C₁e^x + C₂e^2x.

Step 2: Choose the trial solution (y_p).

Since g(x) = 2e^3x, we assume y_p = Ae^3x. Since e^3x is not a term in y_c, no modification is needed.

Step 3: Differentiate and substitute.

y'_p = 3Ae^3x

y''_p = 9Ae^3x

Substitute into the differential equation:

9Ae^3x - 3(3Ae^3x) + 2(Ae^3x) = 2e^3x

(9A - 9A + 2A)e^3x = 2e^3x

2Ae^3x = 2e^3x → 2A = 2 → A = 1.

Step 4: Write the general solution.

y = y_c + y_p = C₁e^x + C₂e^2x + e^3x.

Common Mistakes

• Neglecting Duplication: Forgetting to multiply by x when a trial solution term duplicates a term in y_c. This leads to a mathematical contradiction where terms cancel out to 0.
• Incomplete Sine/Cosine Trial Forms: If g(x) contains only sin(βx), the trial solution must still contain both A cos(βx) and B sin(βx) to account for derivative coupling.

Method of Variation of Parameters

The Method of Variation of Parameters is a general technique to find the particular solution y_p of any linear second-order differential equation. Unlike the method of undetermined coefficients, it does not depend on the form of the non-homogeneous term and works with variable coefficients.

The Formula and Underlying Concept

For a standard second-order linear differential equation:

y'' + P(x)y' + Q(x)y = f(x)

Let the complementary function be y_c = C₁y₁(x) + C₂y₂(x). We vary the parameters (constants C₁ and C₂) by replacing them with functions u₁(x) and u₂(x) to assume a particular solution of the form:

y_p = u₁(x)y₁(x) + u₂(x)y₂(x)

The functions u₁(x) and u₂(x) are determined using the Wronskian of y₁ and y₂:

W(y₁, y₂) = y₁y'₂ - y'₁y₂

The formulas for the variable parameters are:

u₁(x) = - ∫ [ (y₂(x) f(x)) / W(y₁, y₂) ] dx

u₂(x) = ∫ [ (y₁(x) f(x)) / W(y₁, y₂) ] dx

Step-by-Step Solution Procedure

1. Put the differential equation into standard form: y'' + P(x)y' + Q(x)y = f(x) (ensure the coefficient of y'' is 1).
2. Solve the corresponding homogeneous equation to find y_c = C₁y₁ + C₂y₂.
3. Identify y₁ and y₂, then compute the Wronskian W(y₁, y₂).
4. Substitute y₁, y₂, f(x), and W into the formulas for u₁(x) and u₂(x) and integrate.
5. Construct y_p = u₁y₁ + u₂y₂ and write the general solution y = y_c + y_p.

Detailed Example

Solve: y'' + y = sec(x)

Step 1: Solve the homogeneous equation.

y'' + y = 0 → Auxiliary equation: m² + 1 = 0 → m = ±i.

y_c = C₁ cos(x) + C₂ sin(x).

Here, y₁ = cos(x) and y₂ = sin(x).

Step 2: Compute the Wronskian.

W(y₁, y₂) = cos(x)(cos(x)) - (-sin(x))(sin(x)) = cos²(x) + sin²(x) = 1.

Step 3: Compute u₁(x) and u₂(x).

f(x) = sec(x)

u₁(x) = - ∫ [ (sin(x) sec(x)) / 1 ] dx = - ∫ tan(x) dx = ln|cos(x)|

u₂(x) = ∫ [ (cos(x) sec(x)) / 1 ] dx = ∫ 1 dx = x

Step 4: Form the particular and general solutions.

y_p = cos(x) ln|cos(x)| + x sin(x)

y = C₁ cos(x) + C₂ sin(x) + cos(x) ln|cos(x)| + x sin(x).

The Cauchy-Euler Equation and Its Solutions

The Cauchy-Euler equation (also known as the equidimensional equation) is a class of linear differential equations with variable coefficients where the power of the independent variable x matches the order of the derivative.

Definition

A second-order Cauchy-Euler equation is defined as:

ax² d²y/dx² + bx dy/dx + cy = g(x)

where a, b, and c are constants, and x > 0.

Solution Methodology via Transformation

To solve this equation, we use the substitution x = e^z, which implies z = ln(x). Using the chain rule, the derivatives are transformed as:

dy/dx = (1/x) dy/dz

d²y/dx² = (1/x²) [ d²y/dz² - dy/dz ]

Substituting these into the differential equation yields an equation with constant coefficients in terms of z:

a d²y/dz² + (b - a) dy/dz + cy = g(e^z)

The Auxiliary Equation and Cases of Roots

By assuming a solution of the form y = x^m for the homogeneous part ax²y'' + bxy' + cy = 0, we obtain the auxiliary equation:

am(m - 1) + bm + c = 0 → am² + (b - a)m + c = 0

The roots of this auxiliary equation determine the complementary function:

Detailed Example

Solve: x² d²y/dx² - 2x dy/dx - 4y = 0

Step 1: Set up the auxiliary equation.

Here, a = 1, b = -2, c = -4.

The auxiliary equation is:

m(m - 1) - 2m - 4 = 0 → m² - 3m - 4 = 0

Step 2: Solve for the roots.

(m - 4)(m + 1) = 0 → m₁ = 4, m₂ = -1.

Step 3: Construct the general solution.

Since the roots are real and distinct:

y = C₁x⁴ + C₂x^-1

Common Mistakes

• Confusing the Auxiliary Forms: Incorrectly using the auxiliary equation form m² + bm + c = 0 instead of am(m-1) + bm + c = 0. Always remember to subtract 'a' from the linear term's coefficient!

Trajectories: Isogonal and Orthogonal

A trajectory is a curve that cuts every member of a given family of curves at a constant angle. If that angle is a right angle, they are called orthogonal trajectories. If the angle is any other constant, they are called isogonal trajectories.

Orthogonal Trajectories in Cartesian Coordinates

To find the orthogonal trajectories of a family of curves given by F(x, y, c) = 0:

1. Differentiate the equation of the family with respect to x to get an equation containing dy/dx and the constant c.
2. Eliminate the constant c to obtain the differential equation of the family: dy/dx = f(x, y).
3. Since orthogonal curves have perpendicular tangents, their slopes satisfy m₁ × m₂ = -1. Replace dy/dx with -dx/dy.
4. Solve the resulting differential equation: -dx/dy = f(x, y).

Orthogonal Trajectories in Polar Coordinates

For a family of curves given by F(r, θ, c) = 0:

1. Differentiate with respect to θ and eliminate the constant c to get the differential equation of the family in the form: r dθ/dr = f(r, θ).
2. Replace r dθ/dr with -1 / (r dθ/dr), which is equivalent to replacing dr/dθ with -r² dθ/dr.
3. Solve the resulting differential equation to find the orthogonal family.

Detailed Cartesian Example

Find the orthogonal trajectories of the family of parabolas: y = cx².

Step 1: Differentiate and eliminate c.

dy/dx = 2cx

From the original equation, c = y / x². Substitute this back:

dy/dx = 2(y / x²)x → dy/dx = 2y/x.

Step 2: Replace dy/dx with -dx/dy.

-dx/dy = 2y/x

Step 3: Solve the new differential equation.

Separate variables: x dx = -2y dy

Integrate both sides:

∫ x dx = -2 ∫ y dy → x²/2 = -y² + C'

This can be simplified to:

x² + 2y² = C

The orthogonal trajectories are a family of concentric ellipses.

Isogonal Trajectories

Isogonal trajectories intersect the family of curves at a constant angle α ≠ π/2. If the differential equation of the given family is dy/dx = f(x, y), the slope of the trajectory dy/dx = m satisfies the tangent angle relation:

tan(α) = | (m - f(x, y)) / (1 + m f(x, y)) |

Solving this relation for m = dy/dx yields the differential equation of the isogonal trajectories.