Second Order Differential Equations
3.1 Homogeneous Linear Second-Order Differential Equations
A linear second-order differential equation takes the form:
[asciimath]y'' + p(x)y' + q(x)y = f(x)[/asciimath] (3.1.1)
Here, [asciimath]y[/asciimath] is the function we seek, and [asciimath]p(x)[/asciimath], [asciimath]q(x)[/asciimath], and [asciimath]f(x)[/asciimath] are known functions. When referring to non-homogeneous equations, [asciimath]f(x)[/asciimath] is known as the forcing function, representing external forces or influences. We start with the homogeneous case where [asciimath]f(x)=0[/asciimath], and later, we will explore the non-homogeneous case.
[asciimath]y'' + p(x)y' + q(x)y = 0[/asciimath] (3.1.2)
Unique Solution Theorem. If [asciimath]p(x)[/asciimath] and [asciimath]q(x)[/asciimath] are continuous on an open interval [asciimath](a, b)[/asciimath], then the initial value problem has a unique solution within this interval.
Linear Combination Theorem. Suppose [asciimath]y_1(x)[/asciimath] and [asciimath]y_2(x)[/asciimath] are two solutions to the homogeneous Equation 3.1.2 on an open interval [asciimath](a, b)[/asciimath]. Then any linear combination [asciimath]y = c_1y_1 + c_2y_2[/asciimath] is also a solution over the same interval.
The set of solutions,[asciimath]y_1[/asciimath] and [asciimath]y_2[/asciimath], forms a fundamental set or basis for the solution space if they are linearly independent. This implies any solution to Equation 3.1.2 can be expressed as a linear combination of [asciimath]y_1[/asciimath] and [asciimath]y_2[/asciimath]. The Wronskian, W, is crucial in determining their linear independence. For ,[asciimath]y_1[/asciimath] and [asciimath]y_2[/asciimath], the Wronskian at any [asciimath]x_0[/asciimath] in [asciimath](a,b)[/asciimath] must be non-zero to confirm independence:
[asciimath]W = |(y_1(x_0),y_2(x_0)),(y_1'(x_0),y_2'(x_0))|=[/asciimath] [asciimath]y_1(x_0)y'_2(x_0)-y'_1(x_0)y_2(x_0)[/asciimath] (3.1.3)
Theorem on Linear Independence. If [asciimath]p(x)[/asciimath] and [asciimath]q(x)[/asciimath] are continuous on [asciimath](a, b)[/asciimath] and [asciimath]y_1[/asciimath] and [asciimath]y_2[/asciimath] are solutions, then they are linearly independent on [asciimath](a, b)[/asciimath] if and only if the Wronskian W does not equal zero anywhere on [asciimath](a, b)[/asciimath].
Abel’s Theorem. If [asciimath]p(x)[/asciimath] and [asciimath]q(x)[/asciimath] are continuous on [asciimath](a, b)[/asciimath], and [asciimath]x_0[/asciimath] is any point in [asciimath](a, b)[/asciimath], then the Wronskian [asciimath]W(x)[/asciimath] is given by:
[asciimath]W(x) = W(x_0)e^{\int_{x_0}^x p(t) dt}[/asciimath]
Abel’s Theorem is a powerful tool for analyzing the solutions’ behavior across an interval, affirming that if the Wronskian is non-zero at one point and [asciimath]p(x)[/asciimath] is continuous, then the Wronskian remains non-zero across the entire interval.
Equivalence Theorem: For [asciimath]p(x)[/asciimath] and [asciimath]q(x)[/asciimath] continuous on [asciimath](a, b)[/asciimath], and given two solutions [asciimath]y_1[/asciimath] and [asciimath]y_2[/asciimath] of Equation 3.1.2, the following are equivalent:
- The general solution of the equation on [asciimath](a,b)[/asciimath] is [asciimath]y=c_1y_1+c_2y_2[/asciimath]
- [asciimath]{y_1,y_2}[/asciimath] is a fundamental set of solutions of the equation on [asciimath](a,b)[/asciimath]
- [asciimath]{y_1,y_2}[/asciimath] is linearly independent on [asciimath](a,b)[/asciimath]
- The Wronskian of [asciimath]{y_1,y_2}[/asciimath] is nonzero at some point in [asciimath](a,b)[/asciimath]
- The Wronskian of [asciimath]{y_1,y_2}[/asciimath] is nonzero at all points in [asciimath](a,b)[/asciimath]
With these foundational theorems, we have the necessary tools to start solving homogeneous linear second-order differential equations and prepare for the complexities of non-homogeneous cases.
Two solutions to the differential equation [asciimath]y''-11y'+30y = 0[/asciimath] are [asciimath]y_1 = e^(6t)[/asciimath], [asciimath]y_2 = e^(5t)[/asciimath].
a) Find the Wronskian of the solutions and determine if they are linearly independent.
b) Write the general solution to the differential equation.
c) Find the solution satisfying the initial conditions [asciimath]y(0) = -5, \ y'(0) = -32[/asciimath].
Show/Hide Solution
a) To find Wronskian, we use Equation 3.1.3. We first need to find the first derivatives of the solutions [asciimath]y_1[/asciimath] and [asciimath]y_2[/asciimath].
[asciimath]y_1=e^(6t) ->[/asciimath] [asciimath]y_1'=6e^(6t)[/asciimath]
[asciimath]y_2 = e^(5t)->[/asciimath] [asciimath]y'_2 = 5e^(5t)[/asciimath]
[asciimath]W(t) = |(y_1,y_2),(y_1',y_2')|[/asciimath]
[asciimath]W (t)= |(e^(6t),e^(5t) ),(6e^(6t),5e^(5t) ) |[/asciimath]
[asciimath]=-e^(11t)!=0[/asciimath] for any [asciimath]t[/asciimath]
The Wronskian [asciimath]W(t)=-e^(11t)[/asciimath] is never equal to zero for any value of [asciimath]t[/asciimath], which means the solutions are linearly independent on the interval [asciimath](-oo,oo)[/asciimath].
b) Since the solutions are linearly independent, we can express the general solution to the differential equation as a combination of these solutions.
[asciimath]y=c_1y_1+c_2y_2[/asciimath]
[asciimath]y=c_1e^(6t)+c_2e^(5t)[/asciimath]
Here, [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath] are constants that will be determined based on initial conditions or specific requirements of the problem.
c) We apply the initial conditions to find constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
Applying the initial condition to [asciimath]y[/asciimath]:
[asciimath]y(0)=-5[/asciimath]
[asciimath]c_1e^(0)+c_2e^(0)=-5[/asciimath]
[asciimath]c_1+c_2=-5[/asciimath]
Applying the initial condition to [asciimath]y'[/asciimath]:
[asciimath]y'=6c_1e^(6t)+5c_2e^(5t)[/asciimath]
[asciimath]y'(0) = -32[/asciimath]
[asciimath]6c_1e^(0)+5c_2e^(0)=-32[/asciimath]
[asciimath]6c_1+5c_2=-32[/asciimath]
To determine [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath], we need to solve the following system of two equations and two unknowns:
[asciimath]{(c_1+c_2=-5 ),(6c_1+5c_2=-32 ):}[/asciimath]
Solving the system yields
[asciimath]c_1=-7,[/asciimath] [asciimath]c_2=2[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]y=-7e^(6t)+2e^(5t)[/asciimath]
Try an Example
Try an Example
Section 3.1 Exercises
- Compute the Wronskian of the functions [asciimath]y_1 = 2 e^(x/6)[/asciimath] and [asciimath]y_2 = xe^(x/6)[/asciimath]. Determine if the functions are linearly independent for all real numbers.
Show/Hide Answer
[asciimath]W(x)=2 e^(1/3x)[/asciimath] ; the functions are linearly independent because [asciimath]W(x)!=0[/asciimath] for all real numbers.
- Two solutions to the equation [asciimath]y''-y'-2y = 0[/asciimath] are [asciimath]y_1 = e^(-t)[/asciimath], [asciimath]y_2 = e^(2t)[/asciimath].
a) Find the Wronskian.
b) Find the solution satisfying the initial conditions [asciimath]y(0) = 2, \ y'(0) = -11[/asciimath].Show/Hide Answer
a) [asciimath]W(t)=3e^t[/asciimath]
b) [asciimath]y(t)=5 e^(-t)-3 e^(2t)[/asciimath]
- Two solutions to the equation [asciimath]y''+10y'+41y = 0[/asciimath] are [asciimath]y_1 = e^(-5t)sin(4t)[/asciimath], [asciimath]y_2 = e^(-5t)cos(4t)[/asciimath].
a) Find the Wronskian.
b) Find the solution satisfying the initial conditions [asciimath]y(0) = -5, \ y'(0) = 9[/asciimath].Show/Hide Answer
a) [asciimath]W(t)=-4e^(-10t)[/asciimath]
b) [asciimath]y(t)=-4 e^(-5t)sin(4t)-5 e^(-5t)cos(4t)[/asciimath]
