Table of Contents
Last modified on February 14th, 2025
A second-order linear differential equation is an equation of the form:
a(x)y′′ + b(x)y′ + c(x)y = f(x)
Here,
This equation is classified as second-order because it involves the second derivative of an unknown function. It is termed linear as the function y, and its derivatives appear in a linear form (without being multiplied).
Example
A common example is the equation that models the motion of a mass mmm attached to a spring with spring constant k and damping coefficient c:
${m\dfrac{d^{2}x}{dt^{2}}+c\dfrac{dx}{dt}+kx=0}$
This represents simple harmonic motion in a mass-spring system.
Second-order linear differential equations can be classified into two types: homogeneous and non-homogeneous.
A second-order differential equation is homogeneous if:
a(x)y′′ + b(x)y′ + c(x)y = 0
Here,
This means there is no forcing function.
${\dfrac{d^{2}y}{dx^{2}}+4y=0}$ represents a second-order linear differential homogenous equation.
A second-order differential equation is non-homogeneous if:
a(x)y′′ + b(x)y′ + c(x)y = f(x)
Here,
${\dfrac{d^{2}y}{dx^{2}}+2\dfrac{dy}{dx}-y=x}$ represents a second-order linear differential non-homogenous equation.
To solve these equations, we use different methods depending on the type of the equation.
The solution to a homogeneous second-order linear differential equation is obtained using the characteristic equation method.
For a homogeneous equation, we assume a solution of the form: y = erx
Substituting y into the equation gives the characteristic equation:
ar2 + br + c = 0
Solving for r, we obtain the roots that determine the general solution:
Problem: SOLVING a second-order linear differential equation by the CHARACTERISTIC EQUATION METHOD
Solve: y′′ – 7y′ + 12y = 0
Given, y′′ – 7y′ + 12y = 0
Step 1: Forming the Characteristic Equation
r2 – 7r + 12 = 0
Step 2: Solve for r
Factoring the quadratic equation, we get
⇒ r2 – 4r – 3r + 12 = 0
⇒ r(r – 4) – 3(r – 4) = 0
⇒ (r – 3)(r – 4) = 0
⇒ r = 3 and 4
Step 3: Finding the Solution
Since both roots are real and distinct, the solution is:
y = C1 e3x + C2 e4x
Here, C1 and C2 are arbitrary constants.
The solution to a non-homogeneous second-order linear differential equation can be found by the methods of undetermined coefficients or variations of parameters.
If the non-homogeneous term f(x) is a simple function (e.g., polynomial, exponential, sine, or cosine), we assume a trial solution and then determine the coefficients.
In this method, the solution takes the form:
y = yc + yp
Here,
Problem: SOLVING a second-order linear differential equation by the UNDETERMINED COEFFICIENTS METHOD
Solve: y′′ – y = ex
Given, y′′ – y = ex
Step 1: Solving the Homogeneous Equation
y′′ – y = 0
The characteristic equation is:
r2 – 1 = 0
⇒ r = ±1
Thus, the homogeneous solution is:
yc = C1ex + C2e-x
Step 2: Finding a Particular Solution
Since f(x) = ex, we try with:
yp = Axex
Step 3: Differentiating
⇒ ${y_{p}’}$ = Aex + Axex
⇒ ${y_{p}”}$ = Aex + Aex + Axex = 2Aex + Axex
Step 4: Substituting into the Equation
⇒ (2Aex + Axex) – (Axex) = ex
⇒ 2Aex = ex
Step 5: Solving for A
⇒ A = ${\dfrac{1}{2}}$
Thus, the particular solution is:
yp = ${\dfrac{xe^{x}}{2}}$
Step 6: Writing the Final Solution
Thus, the general solution is:
y = ${C_{1}e^{x}+C_{2}e^{-x}+\dfrac{xe^{x}}{2}}$
Here, C1 and C2 are arbitrary constants.
This method is used when g(x) is logarithmic or more complex, and the undetermined coefficients method does not work.
In this method, for the equation: y′′ + ay′ + by = f(x),
The general solution takes the form: y = yc + yp
Here,
For a particular solution of the form yp, we get:
yp = u1y1 + u2y2
⇒ yp = ${-y_{1}\int \dfrac{y_{2}f\left( x\right) }{W}+y_{2}\int \dfrac{y_{1}f\left( x\right) }{W}}$
Here,
Since y1 and y2 are linearly independent, the Wronskian value cannot be equal to zero. ${u_{1}’}$ and ${u_{2}’}$ are derived using the determinant method.
Problem: SOLVING a second-order linear differential equation by the VARIATION OF PARAMETERS METHOD
Solve: ${y”-y=\dfrac{e^{x}}{x}}$
Given, ${y”-y=\dfrac{e^{x}}{x}}$
Step 1: Solving the Homogeneous Equation
The associated homogeneous equation is: y” – y = 0
The characteristic equation is: r2 – 1 = 0 ⇒ r = ±1
Thus, the roots are r1 = 1 and r2 = -1, giving the homogeneous solution:
yc = C1ex + C2e-x
Thus, we have
y1 = ex and y2 = e-x
Step 2: Assuming a Particular Solution
Let us assume yp = u1ex + u2e-x
Differentiating with respect to x,
⇒ ${y_{p}’}$ = ${u_{1}’e^{x}+u_{1}e^{x}+u_{2}’e^{-x}+u_{2}e^{-x}}$
Imposing the simplification condition,
⇒ ${u_{1}’e^{x}+u_{2}’e^{-x}=0}$
Differentiating again, we get
⇒ ${y_{p}”}$ = ${u_{1}’e^{x}+u_{1}e^{x}+u_{2}’e^{-x}+u_{2}e^{-x}}$
Since y′′ – y cancels homogeneous terms, we substitute:
${u_{1}’e^{x}+u_{2}’e^{-x}=0}$ …..(i)
${u_{1}’e^{x}-u_{2}’e^{-x}=\dfrac{e^{x}}{x}}$ …..(ii)
Step 3: Solving for ${u_{1}’}$ and ${u_{2}’}$
Adding equations (i) and (ii),
${2u_{1}’e^{x}=\dfrac{e^{x}}{x}}$
⇒ ${u_{1}’=\dfrac{1}{2x}}$
Subtracting the equations (i) and (ii),
${2u_{2}’e^{-x}=-\dfrac{e^{x}}{x}}$
⇒ ${u_{2}’=-\dfrac{e^{2x}}{2x}}$
Step 4: Integrating to Find u1 and u2
u1 = ${\int \dfrac{1}{2x}dx}$ = ${\dfrac{1}{2}\ln \left| x\right|}$
u2 = ${\int \dfrac{e^{2x}}{2x}dx}$
The second integral is complex and typically solved using integration by parts.
Step 5: Writing the Particular Solution
Thus,
yp = ${\dfrac{1}{2}\left( \ln \left| x\right| \right) e^{x}+\left( \int -\dfrac{e^{2x}}{2x}dx\right) e^{-x}}$
Step 6: Writing the Final Solution
Thus, the general solution is:
y = C1ex + C2e-x + ${\dfrac{1}{2}\left( \ln \left| x\right| \right) e^{x}+\left( \int -\dfrac{e^{2x}}{2x}dx\right) e^{-x}}$
Here, C1 and C2 are arbitrary constants.
Note: Second-order linear differential equations are a specific type of nth-order differential equations, which involve higher-order derivatives. Various solution methods, such as the characteristic equation, undetermined coefficients, and variation of parameters, can be applied to higher-order cases as well. In many instances, these equations can be converted into a system of first-order equations and solved efficiently using matrix methods.
When solving a second-order linear differential equation, the general solution contains two arbitrary constants (denoted as C1 and C2). To obtain a unique solution, we apply initial conditions or boundary conditions that specify particular values for the function and its derivatives at a given point.
Let us now solve a second-order linear differential equation with an initial condition and determine the unique solution.
Solve y′′ – y = 0 with y(0) = 2 and y′(0) = 3.
Given, y” – y’ = 0
Step 1: Solving the Homogeneous Equation
The characteristic equation is:
r2 – 1 = 0 ⇒ r = ±1
Thus, the general solution is: y = C1ex + C2e-x
Step 2: Applying Initial Conditions
y(0) = 2 ⇒ C1e0 + C2e-0 = 2 ⇒ C1 + C2 = 2 …..(i)
y’(x) = C1ex – C2e-x
y’(0) = 3 ⇒ C1e0 – C2e-0 = 3 ⇒ C1 – C2 = 3 …..(ii)
Adding equations (i) and (ii),
(C1 + C2) + (C1 – C2) = 2 + 3
⇒ 2C1 = 5
⇒ C1 = ${\dfrac{5}{2}}$
Subtracting equation (ii) from (i),
(C1 + C2) – (C1 – C2) = 2 – 3
⇒ 2C2 = -1
⇒ C2 = ${-\dfrac{1}{2}}$
Thus, the unique solution is: y = ${\dfrac{5}{2}e^{x}-\dfrac{1}{2}e^{-x}}$
Last modified on February 14th, 2025