Dr. Mohamed Mabrok

Into to DE – 3

Solving Second-Order Linear ODEs

Solving Common Second-Order Linear ODEs

Second-order linear ODEs are fundamental in engineering and physics. We'll cover the primary methods for solving them, focusing on equations with constant coefficients and the Cauchy-Euler equation.

1. Homogeneous Linear Equations with Constant Coefficients

Definition

Equations of the form:

\[ ay'' + by' + cy = 0 \]

where \(a, b, c\) are real constants and \(a \neq 0\).

Methodology (Characteristic Equation)

  1. Assume a solution of the form \( y = e^{rx} \). Substitute this into the ODE: \( a(r^2 e^{rx}) + b(r e^{rx}) + c(e^{rx}) = 0 \).
  2. Since \( e^{rx} \) is never zero, divide by it to get the characteristic (or auxiliary) equation: \( ar^2 + br + c = 0 \).
  3. Solve this quadratic equation for \(r\). There are three cases based on the discriminant \( \Delta = b^2 - 4ac \):
  4. Case 1: Distinct Real Roots (\(\Delta > 0\))
    Two distinct real roots \(r_1, r_2\).
    The general solution is: \( y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x} \)
  5. Case 2: Repeated Real Root (\(\Delta = 0\))
    One real root \(r = -b/2a\).
    The general solution is: \( y(x) = (c_1 + c_2 x) e^{r x} \)
  6. Case 3: Complex Conjugate Roots (\(\Delta < 0\))
    Two complex roots \(r = \alpha \pm i\beta\), where \(\alpha = -b/2a\) and \(\beta = \sqrt{4ac - b^2}/2a\).
    The general solution is: \( y(x) = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x)) \)

Examples

a) Solve: \( y'' + 5y' + 6y = 0 \)

Characteristic Eq: \( r^2 + 5r + 6 = 0 \Rightarrow (r+2)(r+3) = 0 \). Roots \( r_1 = -2, r_2 = -3 \) (Distinct Real).
Solution: \( y(x) = c_1 e^{-2x} + c_2 e^{-3x} \)

b) Solve: \( y'' - 6y' + 9y = 0 \)

Characteristic Eq: \( r^2 - 6r + 9 = 0 \Rightarrow (r-3)^2 = 0 \). Root \( r = 3 \) (Repeated Real).
Solution: \( y(x) = (c_1 + c_2 x) e^{3x} \)

c) Solve: \( y'' + 4y' + 13y = 0 \)

Characteristic Eq: \( r^2 + 4r + 13 = 0 \). Using quadratic formula: \( r = \frac{-4 \pm \sqrt{16 - 4(1)(13)}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i \). Roots are complex with \( \alpha = -2, \beta = 3 \).
Solution: \( y(x) = e^{-2x} (c_1 \cos(3x) + c_2 \sin(3x)) \)

2. Non-Homogeneous Linear Equations with Constant Coefficients

Definition

Equations of the form:

\[ ay'' + by' + cy = g(x) \]

where \(a, b, c\) are constants (\(a \neq 0\)) and \( g(x) \) is a non-zero function.

The general solution is \( y(x) = y_c(x) + y_p(x) \), where \( y_c \) is the solution to the homogeneous equation \( ay'' + by' + cy = 0 \), and \( y_p \) is any particular solution.

General Methodology

  1. Find the complementary function \(y_c(x)\) by solving the associated homogeneous equation \( ay'' + by' + cy = 0 \) using the characteristic equation method (Section 1).
  2. Find a particular solution \(y_p(x)\) using one of the methods below (Undetermined Coefficients or Variation of Parameters).
  3. The general solution is the sum: \( y(x) = y_c(x) + y_p(x) \).

Method 1: Undetermined Coefficients

This method works when \(g(x)\) is a polynomial, exponential (\(e^{kx}\)), sine (\(\sin(kx)\)), cosine (\(\cos(kx)\)), or finite sums and products of these functions.

  1. Based on the form of \(g(x)\), make an educated guess for the form of \(y_p(x)\) with unknown coefficients.
    • If \(g(x)\) is polynomial of degree \(n\), guess \(y_p\) as a full polynomial of degree \(n\).
    • If \(g(x) = A e^{kx}\), guess \(y_p = B e^{kx}\).
    • If \(g(x) = A \cos(kx)\) or \(A \sin(kx)\), guess \(y_p = B \cos(kx) + D \sin(kx)\).
    • Combine these for sums/products (e.g., if \(g(x)=x e^{2x}\), guess \(y_p=(Ax+B)e^{2x}\)).
  2. Modification Rule: Compare your initial guess for \(y_p\) with the terms in \(y_c\). If any term in the guess is already present in \(y_c\), multiply the entire guess by \(x\). If it's still present, multiply by \(x\) again.
  3. Substitute the (modified) guess for \(y_p\) and its derivatives into the non-homogeneous ODE.
  4. Equate coefficients of like terms on both sides to solve for the unknown coefficients in your guess.
  5. Substitute the found coefficients back into \(y_p\).
Example (Undetermined Coefficients)

Solve: \( y'' - 2y' - 3y = 5e^{3x} \)

1. Find \(y_c\): Char. Eq: \( r^2 - 2r - 3 = 0 \Rightarrow (r-3)(r+1) = 0 \). Roots \( r_1=3, r_2=-1 \). \( y_c = c_1 e^{3x} + c_2 e^{-x} \).
2. Find \(y_p\): \(g(x) = 5e^{3x}\). Initial guess \( y_p = A e^{3x} \).
3. Modification Rule: The guess \(A e^{3x}\) is part of \(y_c\). Multiply by \(x\): New guess \( y_p = Axe^{3x} \).
4. Derivatives: \( y_p' = A(e^{3x} + 3xe^{3x}) \), \( y_p'' = A(3e^{3x} + 3e^{3x} + 9xe^{3x}) = A(6e^{3x} + 9xe^{3x}) \).
5. Substitute into ODE: \( A(6e^{3x} + 9xe^{3x}) - 2A(e^{3x} + 3xe^{3x}) - 3(Axe^{3x}) = 5e^{3x} \).
6. Simplify: \( e^{3x} [ (9A - 6A - 3A)x + (6A - 2A) ] = 5e^{3x} \).
\( \implies e^{3x} [ 0x + 4A ] = 5e^{3x} \). So \( 4A = 5 \Rightarrow A = 5/4 \).
7. Particular solution: \( y_p = \frac{5}{4}xe^{3x} \).
8. General Solution: \( y = y_c + y_p \).
Solution: \( y(x) = c_1 e^{3x} + c_2 e^{-x} + \frac{5}{4}xe^{3x} \)

Method 2: Variation of Parameters

This method works for any function \(g(x)\), provided you know the complementary function \( y_c = c_1 y_1(x) + c_2 y_2(x) \). It's generally more complex computationally than Undetermined Coefficients.

  1. Find the complementary function \( y_c = c_1 y_1(x) + c_2 y_2(x) \).
  2. Assume the particular solution has the form \( y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x) \).
  3. Calculate the Wronskian: \( W(y_1, y_2) = y_1 y_2' - y_1' y_2 \).
  4. Find \( u_1'(x) \) and \( u_2'(x) \) using the formulas: \[ u_1'(x) = - \frac{y_2(x) g(x)}{a W(y_1, y_2)} \] \[ u_2'(x) = \frac{y_1(x) g(x)}{a W(y_1, y_2)} \] (Remember to include the leading coefficient \(a\) from \(ay''+...\) if it's not 1).
  5. Integrate \( u_1'(x) \) and \( u_2'(x) \) to find \( u_1(x) \) and \( u_2(x) \). (No need for constants of integration here, they get absorbed into \(y_c\)).
  6. Construct the particular solution: \( y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x) \).
Example (Variation of Parameters)

Solve: \( y'' + 4y = \sec(2x) \)

1. Find \(y_c\): Char. Eq: \( r^2 + 4 = 0 \Rightarrow r = \pm 2i \). \( y_c = c_1 \cos(2x) + c_2 \sin(2x) \). So \( y_1 = \cos(2x), y_2 = \sin(2x) \). Here \(a=1\).
2. Calculate Wronskian: \( y_1' = -2\sin(2x) \), \( y_2' = 2\cos(2x) \). \( W = (\cos(2x))(2\cos(2x)) - (-2\sin(2x))(\sin(2x)) = 2\cos^2(2x) + 2\sin^2(2x) = 2 \).
3. Find \(u_1', u_2'\): \( g(x) = \sec(2x) = 1/\cos(2x) \). \[ u_1' = - \frac{\sin(2x) \sec(2x)}{1 \cdot W} = - \frac{\sin(2x) (1/\cos(2x))}{2} = -\frac{1}{2}\tan(2x) \] \[ u_2' = \frac{\cos(2x) \sec(2x)}{1 \cdot W} = \frac{\cos(2x) (1/\cos(2x))}{2} = \frac{1}{2} \]
4. Integrate: \( u_1 = \int -\frac{1}{2}\tan(2x) dx = -\frac{1}{2} \left( \frac{1}{2}\ln|\sec(2x)| \right) = \frac{1}{4}\ln|\cos(2x)| \) (using \(\int \tan(u) du = -\ln|\cos u|\) or \(\ln|\sec u|\)). \( u_2 = \int \frac{1}{2} dx = \frac{1}{2}x \).
5. Construct \(y_p\): \( y_p = u_1 y_1 + u_2 y_2 = \left( \frac{1}{4}\ln|\cos(2x)| \right) \cos(2x) + \left( \frac{1}{2}x \right) \sin(2x) \).
6. General Solution: \( y = y_c + y_p \).
Solution: \( y(x) = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{4}\cos(2x)\ln|\cos(2x)| + \frac{1}{2}x\sin(2x) \)

3. Cauchy-Euler Equations

Definition

Equations with variable coefficients of the specific form:

\[ ax^2 y'' + bxy' + cy = g(x) \]

where \(a, b, c\) are constants (\(a \neq 0\)). Typically solved for \(x > 0\).

Methodology (Homogeneous Case)

  1. Assume a solution of the form \( y = x^r \). Calculate \( y' = rx^{r-1} \) and \( y'' = r(r-1)x^{r-2} \).
  2. Substitute into the homogeneous equation \( ax^2 y'' + bxy' + cy = 0 \): \( ax^2(r(r-1)x^{r-2}) + bx(rx^{r-1}) + c(x^r) = 0 \).
  3. Simplify and factor out \( x^r \) (since \(x > 0\)): \( [ar(r-1) + br + c] x^r = 0 \).
  4. Solve the auxiliary equation: \( ar(r-1) + br + c = 0 \) or \( ar^2 + (b-a)r + c = 0 \).
  5. Find the roots \(r\). There are three cases:
  6. Case 1: Distinct Real Roots \(r_1, r_2\)
    General solution: \( y_c(x) = c_1 x^{r_1} + c_2 x^{r_2} \)
  7. Case 2: Repeated Real Root \(r\)
    General solution: \( y_c(x) = (c_1 + c_2 \ln x) x^r \)
  8. Case 3: Complex Conjugate Roots \(r = \alpha \pm i\beta\)
    General solution: \( y_c(x) = x^{\alpha} [c_1 \cos(\beta \ln x) + c_2 \sin(\beta \ln x)] \)
  9. For the non-homogeneous case \( ax^2 y'' + bxy' + cy = g(x) \), find \( y_c \) as above, then use Variation of Parameters to find \( y_p \). Remember to use \(g(x)/a\) in the formulas if \(a \neq 1\).

Alternative: Use the substitution \(x=e^t\) (\(t=\ln x\)) to transform it into a constant coefficient equation in \(t\).

Example (Homogeneous Cauchy-Euler)

Solve: \( x^2y'' + 5xy' + 4y = 0 \) for \( x > 0 \)

1. Auxiliary Eq: \( a=1, b=5, c=4 \). \( r(r-1) + 5r + 4 = 0 \).
2. Simplify: \( r^2 - r + 5r + 4 = 0 \Rightarrow r^2 + 4r + 4 = 0 \).
3. Solve: \( (r+2)^2 = 0 \). Root \( r = -2 \) (Repeated Real).
Solution: \( y(x) = (c_1 + c_2 \ln x) x^{-2} \)

4. Reduction of Order

Definition

A technique used to find a second, linearly independent solution \(y_2(x)\) to a second-order linear homogeneous ODE \(a(x)y'' + b(x)y' + c(x)y = 0\), given that one non-trivial solution \(y_1(x)\) is already known.

Methodology

  1. Start with the homogeneous ODE \(a(x)y'' + b(x)y' + c(x)y = 0\) and known solution \(y_1(x)\).
  2. Assume the second solution is of the form \( y_2(x) = v(x) y_1(x) \).
  3. Calculate derivatives: \( y_2' = v'y_1 + vy_1' \) and \( y_2'' = v''y_1 + 2v'y_1' + vy_1'' \).
  4. Substitute \( y_2, y_2', y_2'' \) into the ODE.
  5. Simplify the equation. The terms involving \(v\) (without derivatives) will cancel out because \(y_1\) is a solution. This leaves an equation involving \(v''\) and \(v'\).
  6. Let \( w = v' \). The equation becomes a first-order linear ODE in \(w\). Solve for \(w\).
  7. Integrate \(w\) to find \(v\): \( v = \int w dx \).
  8. The second solution is \( y_2 = v y_1 \). (Any constant multiple can be ignored).
  9. The general solution is \( y(x) = c_1 y_1(x) + c_2 y_2(x) \).
  10. Alternatively, use the formula derived from this process (ensure ODE is in form \(y''+P(x)y'+Q(x)y=0\)): \[ y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{(y_1(x))^2} dx \]

Example (Reduction of Order)

Given that \( y_1(x) = x \) is a solution to \( x^2y'' - x(x+2)y' + (x+2)y = 0 \) for \( x > 0 \), find the general solution.

1. Standard form: \( y'' - \frac{x+2}{x}y' + \frac{x+2}{x^2}y = 0 \). So \( P(x) = -\frac{x+2}{x} = -1 - \frac{2}{x} \).
2. Calculate exponent term for formula: \( -\int P(x) dx = -\int (-1 - 2/x) dx = \int (1 + 2/x) dx = x + 2\ln x = x + \ln(x^2) \). So \( e^{-\int P(x) dx} = e^{x + \ln(x^2)} = e^x e^{\ln(x^2)} = x^2 e^x \).
3. Apply formula: \( y_2 = y_1 \int \frac{e^{-\int P(x) dx}}{(y_1)^2} dx = x \int \frac{x^2 e^x}{(x)^2} dx = x \int e^x dx \).
4. Integrate: \( y_2 = x (e^x) = xe^x \).
General Solution: \( y(x) = c_1 x + c_2 xe^x \)