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Solution of a Second-Order Linear Homogeneous ODE

Mathematics is the language in which God has written the universe, Galileo Galilei

Mathematics is not about numbers, equations, computations, or algorithms: it is about understanding, William Paul Thurston

Recall

Differential equations

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

  1. Dependent and independent variables. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.
  2. Constants. Fixed numerical values that do not change.
  3. Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x0, y0) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

Second-order Linear Homogeneous ODE’s

Second-order Linear Homogeneous ODE’s

Second-order linear homogeneous ordinary differential equations (ODEs) are fundamental in mathematical modeling and appear frequently in physics, engineering, and other sciences. They describe a wide range of phenomena, such as mechanical vibrations, electrical circuits, and fluid dynamics.

Definition. A second-order linear homogeneous ODE is a differential equation of the form: y'' + p(x)y' + q(x)y = 0 where:

The term linear refers to the fact that y, y’, and y′′ appear to the first power only -there are no squares (y2), cubes (y’(3)), or other nonlinear expression.

The term homogeneous means the right-hand side of the equation is zero. In other words, the equation is set equal to zero, indicating that there are no external forcing functions.

General Solution of a Second-Order Linear Homogeneous ODE

The goal when solving such ODEs is to find the general solution, which represents all possible solutions.

The general solution to a second-order linear homogeneous ODEs is: y = c1y1 + c2y2 where

Importance of Linear Independence

Two solutions y1 and y2 are said to be linearly independent if neither is a constant multiple of the other. That is, there is no constant c or c’ such that y2 ≠ cy1 and y1 ≠ c’y2 over the interval of interest.

Why is Linear Independence Important?

Test for Linear Independence

One common method to test for linear independence is using the Wronskian determinant: $W(y_1, y_2)(x) = y_1(x)y_2'(x) -y_1'(x)y_2(x)$. If W(y1, y2)(x) ≠ 0 for some x in the interval, then y1 and y2 are linearly independent.

The superposition principle applies to linear homogeneous differential equations and states: If y1(x) and y2(x) are solutions to a linear homogeneous ODE, then any linear combination of y1 and y2, i.e., y = c1y1 + c2y2 is also a solution

It allows us to construct a general solution from known solutions and underpins the structure of the solution space for linear homogeneous ODEs.

Proof 1. Direct Substitution

We will present two proofs to demonstrate why linear combinations of solutions are also solutions.

Consider the ODE: y’’ + p(x)y’ + q(x)y = 0. Suppose y1 and y2 are known solutions to the ODE, meaning:

y1’’ + p(x)y1’ + q(x)y1 = 0
and
y2’’ + p(x)y2’ + q(x)y2 = 0

Now, take a linear combination y = c1y1 + c2y2, where c1 and c2 are constant. Let’s verify that this y is also a solution.

First derivative: y’ = c1y’1 + c2y’2. Second derivative: c1y’’1 + c2y’’2

Next, substitute these into the original equation y’’ + p(x)y’ + q(x)y = 0:

(c1y’’1 + c2y’’2) + p(x)(c1y’1 + c2y’2) + q(x)(c1y1 + c2y2) = 0 ↭[🚀]

By distributing c1 and c2, we get:

↭[🚀] c1(y’’1 + p(x)y’1 + y1) + c2(y’’2 + p(x)y’2 + y2) = 0 ↭[🚀]

Since y1 and y2 are both solutions: y’’1 + p(x)y’1 + y1 = 0 and y’’2 + p(x)y’2 + y2 = 0. Therefore,

↭[🚀] c1·0 + c2· 0 = 0. Thus, y = c1y1 + c2y2 satisfies the original equation, and is indeed a solution ∎

Proof 2. Using Linear Operators

Our linear homogeneous ODE is normally written as y’’ + p(x)y’ + q(x)y = 0. We are going to rewrite it using the Differentiator operator as D2y + pDy + qy = 0, then factor y: (D2 + pD + q)·y = 0.

Now, we are going to consider D2 + pD + q as a linear operator and abbreviate it as L, so our equation is simplified to L·y = 0 where L = D2 + pD + q ⇒ L(y) = y’’ + p(x)y’ + q(x)y. L is kind of a black box or function with inputs u(x) and outputs v(x), the result of applying the operator L to u(x).

L is a linear operator, meaning it satisfies two important properties:

  1. Additivity. The operator applied to the sum of two functions is the same as the operator applied to each function separately and then added together: L(u1 + u2) = L(u1) + L(u2).
  2. Homogeneity (scaling). The operator applied to a constant multiple of a function is the same as the constant times the operator applied to the function: L(cu) = cL(u)

where c is a constant, u, u1, and u2 are functions.

For example, the differential operator is a linear operator: (u1 + u2)’ = u1’ + u2’ and (cu)’ = cu’.

Additivity. Suppose u1 and u2 are two functions. Then, the operator L applied to the sum of u1 and u2 is:

L(u1 + u2) = (u1 + u2)’’ + p(x)(u1 + u2)’ + q(x)(u1 + u2)

Using the basic properties of derivatives and addition, we can split each term:

L(u1 + u2) = u1’’ + u2’’ +p(x)(u1’ + u2’) + q(x)(u1 + u2)

Now, rearrange terms:

L(u1 + u2) = (u1’’ + p(x)u1’ + q(x)u1) + (u2’’ + p(x)u2’ + q(x)u2) = L(u1) + L(u2).

Homogeneity (Scaling). Next, we need to show that L satisfies the scaling property. Suppose c is a constant and u is a function. Then applying L to the constant multiple c⋅u gives: L(c·u) = (c⋅u)′′ +p(x)(c⋅u)′ +q(x)(c⋅u)

Since the derivative of a constant multiple of a function is just the constant times the derivative of the function, we can factor out the constant c from each term:

L(c⋅u) = c⋅u′′ +p(x)⋅c⋅u′ +q(x)⋅c⋅u =[Factor c out] = c·(u’’ + p(x)u’ + q(x)) = c·L(u).

Our ODE is expressed as L·y = 0. Because L is a linear operator (it satisfies both additivity and homogeneity), it follows that the superposition principle holds, that is, if y1 and y2 are solutions to the ODE L·y = 0 or L(y) = 0, then any linear combination of these solutions, i.e., c1y1 + c2y2, will also be a solution.

L·(c1y1 + c2y2) =[Additivity] L(c1y1) + L(c2y2) =[Homogeneity (Scaling)] c1L(y1) + c2L(y2) =[Since both y1 and y2 are solutions to the ODE] c1·0 + c2·0 = 0. Thus, y = c1y1 +c2y2 is also a solution.

Why Is This the Complete Family of Solutions?

For second-order linear homogeneous ODEs:

Examples

Characteristic Equation: r2 −5r +6 = 0. It roots are: r=2, r=3

Second-order linear ODE’s. Real and distinct roots. In this case, the solution is a combination of two exponentially terms. The general solution is y(x) = $c_1e^{2x}+c_2e^{3x}$

  1. e2x and e3x are linearly independent solutions.
  2. Constants c1 and c2 are determined by initial conditions.

This is Euler’s Equation.

Assumed Solution: y(x) = xr

Substitute into the ODE: x2(r(r−1)xr−2) −x(rxr−1) +xr = 0 ↭ r(r−1)xr −rxr +xr = 0

Simplify: [r(r-1) -r + 1]xr = 0 ⇒ [r2 -2r + 1]xr = 0.    Characteristic Equation: r2 −2r + 1 = 0. Factoring this, we find: (r−1)2 = 0. The roots of the characteristic equation are: r=1 (repeated root).

  1. When the characteristic equation has distinct roots r1 and r2, the general solution is: $y(x) = c_1x^{r_1}+c_2x^{r_2}$
  2. For constant coefficient differential equations, when we have repeated roots r = r0, the general solution is: y(t) = c1er0t + c2ter0t.
  3. Given one solution y1(x) = xr with r = 1, we assume the second solution is of the form y2(x) = y1(x)·v(x) = xrv(x).

We need to find v(x) such that y2(x) is a solution to the original differential equation.

Compute derivatives: y2’(x) = rxr-1v(x) + xrv’(x)

y2’’(x) = r(r-1)xr-2v(x) + rxr-1v’(x) + rxr-1v’(x) + xrv’’(x) = r(r-1)xr-2v(x) + 2rxr-1v’(x) + xrv’’(x)

Substitute y2 into the differential equation x2y′′ −xy′ +y = 0: x2(r(r-1)xr-2v(x) + 2rxr-1v’(x) + xrv’’(x)) −x(rxr-1v(x) + xrv’(x)) + xrv(x) = r(r-1)xrv(x) + 2rxr+1v’(x) + xr+2v’’(x) -rxrv(x) -xr+1v’(x) + xrv(x) = [Combine like terms] xr+2v’’(x) + (2r-1)xr+1v’(x) + (r(r-1)-r+1)v(x) =[Given r = 1] x3v’’(x) + (2-1)x1+1v’(x) + (0-1+1)v(x) = 0 ⇒ x3v’’(x) + x2v’(x) = 0 ⇒[Divide the entire equation by x2 (since x > 0)] xv’’(x) + v’(x) = 0

This is a first-order linear ODE for v′(x). Let u = v′(x): xu’(x) + u(x) = 0⇒[This can be rewritten as:] $u’(x) + \frac{1}{x}u(x) = 0$.

This is a separable equation: $\frac{du}{u} = \frac{-1}{x}dx⇒[\text{Integrate both sides}] ln|u| = -ln|x| + C ↭ u = \frac{K}{x}$ where K = eC is a constant. Thus, $v’(x) = \frac{K}{x} ⇒ v(x) = Kln(x) + C’$. Since C’ is an arbitrary constant, we can set K = 1 without loss of generality (absorbing constants into C’). Therefore, the second solution is y2(x) = xrv(x) = x1·ln(x).

For two functions y1(x) and y2(x) to be linearly independent, the Wronskian W(y1, y2)(x) must be non-zero.

y1(x) = x, y1’(x) = 1.y’2(x) = $x\frac{1}{x}+ln(x) = 1 + ln(x)$

Since W(y1, y2)(x) = $y_1(x)y_2’(x)-y_1’(x)y_2(x) = x(ln(x)+1)-1·xln(x) = x ≠ 0$ for x > 0, the solutions are linearly independent.

General Solution: y(x) = c1xr + c2xrln(x) =[Substituting r=1:] c1x + c2xln(x) where:

Given initial conditions y(x0)= a, y’(x0) = b. the general solution y(x) = c1x + c2xln(x) and its derivative y’(x) = c1 + c2(ln(x) + 1).

$\begin{cases} c_1x_0 + c_2x_0ln(x_0) = a \\ c_1 + c_2(ln(x_0)+1) = b \end{cases}$

This system can be solved for c1 and c2 provided that the Wronskian(y1, y2)(x0) = x0 ≠ 0, ensuring a unique solution exists.

Solving the initial value problem and Uniqueness

Let y1(x) and y2(x) be two linearly independent solutions to the differential equation y′′ +p(x)y′ +q(x)y = 0

Proposition. The family of solutions {c1y1 + c2y2} is sufficient to satisfy any initial conditions for the second-order linear homogeneous differential equation y′′ +p(x)y′ +q(x)y = 0.

Proof.

Let y(x0) = a, y’(x0) = b be the initial values at some point x0. The general solution to the homogeneous equation is of the form: y(x) = c1y1(x) + c2y2(x) where y1(x) and y2(x) are two independent solutions to the differential equation.

Taking the derivative of the general solution y(x), we get: y’ = c1y1’ + c2y2’. Substitute the initial conditions y(x0)=a and y′(x0)=b into these expressions:

$\begin{cases} c_1y_1(x_0) + c_2y_2(x_0) = a \\ c_1y_1’(x_0) + c_2y_2’(x_0) = b \end{cases}$

This forms a system of two linear equations in the unknowns (variables) c1 and c2. For the system to have a unique solution for c1 and c2, the determinant of the coefficient matrix, which is called the Wronskian, must be non-zero at x0:

$W(y_2, y_2)(x_0) = \Bigl \vert\begin{smallmatrix}y_1(x_0) & y_2(x_0)\\ y_1’(x_0) & y_2’(x_0)\end{smallmatrix}\Bigr \vert ≠ 0$

The Wronskian of two differentiable functions f and g is defined as W(f, g) = fg’ -gf’. More generally, W(f1, ···, fn) = $\Biggl \vert\begin{smallmatrix}f_1(x) & f_2(x) & ··· & f_n(x)\\ f_1’(x) & f_2’(x) & ··· & f_n’(x)\\· & · & ··· & ·\\· & · & ··· & ·\\· & · & ··· & ·\\f_1^{n-1}(x) & f_2^{n-1}(x) & ··· & f_n^{n-1}(x)\end{smallmatrix}\Biggr \vert$. This is the determinant of the matrix constructed by placing the functions in the first row, the first derivatives of the functions in the second row, and so on. In our particular case, W(y1, y2) = $\Bigl \vert\begin{smallmatrix}y_1 & y_2\\ y_1’ & y_2’\end{smallmatrix}\Bigr \vert = y_1(x)y_2’(x)-y_1’(x)y_2(x)$

This determinant plays a crucial role in determining whether y1 and y2 are linearly independent. If W(y1, y2) ≠ 0 for some x, the functions y1 and y2 are linearly independent, meaning that they form a valid basis for the space of solutions to the differential equation. If W(y1, y2) = 0 for all x, the two functions are linearly dependent, which means one is a constant multiple of the other.

Conclusion: Because the Wronskian W(y1, y2)(x0) ≠ 0, we can solve the system of equations for c1 and c2, ensuring that the general solution satisfies the given initial conditions. Thus, the family {c1y1(x) + c3y2(x)} is sufficient to satisfy any initial conditions ∎

Theorem on the Wronskian. If y1 and y2 are solutions to the linear homogeneous ODE y′′ +p(x)y′ +q(x)y = 0, and p(x) and q(x) are continuous on an interval I, then the Wronskian W(y1, y2) either vanishes identically (i.e., W(y1, y2) ≡ 0, W(y1, y2)(x) = 0 ∀x ∈ I) or is never zero on I (i.e., W(y1, y2)(x) ≠ 0 ∀x ∈ I).

Implications

Completeness of the Solution Set

Claim: The set of solutions {c1y1 + c2y2} forms the complete family of solutions to the differential equation. Note: However, y1 and y2 are not necessarily the only independent solutions (“the only game in town”). It is possible that other pairs of independent solutions, say u1(x) and u2(x), can also form a valid basis for the solution space. In this case, the general solution would still be expressed as: {c1u1 + c2u2}

Relationship Between Different Solution Pairs:

However, any other pair u1 and u2 can be expressed as linear combinations of y1 and y2, meaning there is a relationship:

$\begin{cases} u_1 = \bar {c_1}y_1 + \bar {c_2}y_2 \\ u_2 = \bar {\bar {c_1}}y_1 + \bar {\bar {c_2}}y_1 \end{cases}$

We can also define “normalized solutions”, say Y1 and Y2, which are particular solutions of the ODE that satisfy specific initial conditions at x0 (often x0 = 0): Y1(0) = 1, Y1’(0) = 0, and Y2(0) = 0, Y2’(0) = 1.

Purpose of Normalized Solutions: convenience (they simplify the process of solving initial value problems) and direct application (Any solution satisfying initial conditions can be directly expressed as a linear combination of Y1(x) and Y2(x)).

Standard solutions y1 = cos(x), y2(x) = sin(x).

Normalized solutions Y1(x) = cos(x) (since cos(0) = 1 and cos’(0) = 0), = Y2(x) = sin(x) (since sin(0) = 1 and sin’(0) = 0).

Standard solutions are y1 = ex, y2(x) = e-x and the general solution is y = c1ex + c2e-x, y’ = c1ex - c2e-x

Find Normalized Solutions. To find Y1(x):

$\begin{cases}c_1e^0 + c_2e^0 = c_1 + c_2 = 1 \\ c_1e^0 - c_2e^0 = c_1 -c_2 = 0 \end{cases}$

The solution is $c_1 = c_2 = \frac{1}{2}, Y_1 = \frac{e^x+e^{-x}}{2} = cosh(x)$

To find Y2:

$\begin{cases} c_1e^0 + c_2e^0 = c_1 + c_2 = 0 \\ c_1e^0 - c_2e^0 = c_1 -c_2 = 1 \end{cases}$

The solution is $c_1 = \frac{1}{2}, c_2 = \frac{-1}{2}, Y_2 = \frac{e^x-e^{-x}}{2} = sinh(x)$

Advantages of Normalized Solutions

If we have two normalized solutions, say Y1 and Y2, at the point x = 0, then the the solution to the Initial Value Problem (IVP): ODE + $\begin{cases} y(0) = a = y_0 \\ y’(0) = b = y’_0 \end{cases}$ is: y = aY1 + bY2

This works because the initial conditions are satisfied as follows: y(0) = aY1(0) + bY2(0) = a·1 + b·0 = a, y’(0) = aY’1(0) + b’Y2(0) = a·0 + b·1= b

Existence and Uniqueness Theorem for Differential Equations

It states that for a second-order linear homogeneous differential equation of the form: y’’ + py’ + qy = 0 where p(x) and q(x) are continuos functions on an interval containing x0, there exists exactly one solution (both existence and uniqueness) that satisfies the given initial conditions y(x0) = a, and y'(x0) = b.

This means that the family of solutions {c1Y1 + c2Y2} encompasses all possible solutions to the differential equation. Any solution u(x) with initial conditions u(0) = u0, u'(0) = u'0, can be written as: u0Y1 + u'0Y2.

Thus, all solutions to the differential equation belong to the family {c1Y1 + c2Y2}, and if another solution v(x) satisfies the same initial conditions v(0) = u0, v’(0) = u’0, by the uniqueness theorem, u(x) = v(x).

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
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  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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  5. Michael Penn, and MathMajor.
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  8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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