Second Order Differential Equations
3.2 Constant Coefficients Homogeneous Equations
We first consider the homogenous equation with constant coefficients:
[asciimath]a y'' + by' + cy = 0[/asciimath] (3.2.1)
To solve this, we recognize that a solution to this equation must have the property that its second derivative can be expressed as a linear combination of the first derivative and the function itself, suggesting that the solution form is [asciimath]y=e^(rx)[/asciimath]. Substituting [asciimath]y=e^(rx)[/asciimath] and its derivatives into Equation 3.2.1 leads to
[asciimath]e^(rx)(ar^2+br+c)=0[/asciimath]
Since [asciimath]e^(rx)[/asciimath] is never zero for any real number [asciimath]x[/asciimath], we can conclude
[asciimath]ar^2+br+c=0[/asciimath] (3.2.2)
Equation 3.2.2 is known as the auxiliary equation or characteristic equation (characteristic polynomial) of the homogeneous Equation 3.2.1. To determine the general solution of Equation 3.2.1, we solve for [asciimath]r[/asciimath] in the characteristic equation.
The roots of the characteristic equation determine the nature of the solution, leading to three possible cases based on whether the roots are real and distinct, real and repeated, or complex conjugate.
The General Solution to the Second-Order Linear DE with Constant Coefficients
Case 1: Two Distinct Real Roots
If the characteristic equation (Equation 3.2.2) has two real roots [asciimath]r_1[/asciimath] and [asciimath]r_2[/asciimath] , then the solutions are [asciimath]y_1=e^(r_1x)[/asciimath] and [asciimath]y_2=e^(r_2x)[/asciimath]. The general solution is the linear combination of these two solutions:
[asciimath]y=c_1 e^(r_1 x) + c_2 e^(r_2 x)[/asciimath]
Case 2: Repeated Root
If the characteristic equation has a repeated root [asciimath]r[/asciimath], then the solutions are [asciimath]y_1=e^(rx)[/asciimath] and [asciimath]y_2=xe^(rx)[/asciimath]. The general solution is the linear combination of these two solutions:
[asciimath]y=c_1 e^(r x) + c_2 xe^(r x)[/asciimath]
Case 3: Complex Conjugate Roots
If the characteristic equation has complex conjugate roots of the form [asciimath]r=alpha +-ibeta[/asciimath], then the solutions can be represented using Euler’s formula as [asciimath]y_1,y_2=e^(alphax)(cos(betax)+-i sin (betax))[/asciimath]. The real-valued general solution derived from these complex solutions is
[asciimath]y=e^(alphax)(c_1cos(betax)+c_2sin (betax))[/asciimath]
In this form, [asciimath]e^(alpha x)[/asciimath] represents the exponential growth or decay, and the combination of cosine and sine functions represents the oscillatory behavior due to the complex part of the roots.
Find the general solution to the differential equation
[asciimath]y'' + y' - 6y = 0[/asciimath]
Show/Hide Solution
The auxiliary equation is
[asciimath]r^2+r-6=0[/asciimath]
The equation is factorable to
[asciimath](r+3)(r-2)=0[/asciimath]
The roots are [asciimath]r_1=-3[/asciimath] and [asciimath]r_2=2[/asciimath] . This is Case 1 since the roots are real and distinct. Therefore, the general solution is the linear combination of [asciimath]y_1=e^(-3x)[/asciimath] and [asciimath]y_2=e^(2x)[/asciimath]:
[asciimath]y(x)=c_1e^(-3x)+c_2e^(2x)[/asciimath]
Solve the following initial value problem (IVP).
[asciimath]y'' -8 y' +15y = 0,[/asciimath] [asciimath]y(0) = 9, \quad y'(0) = 35[/asciimath]
Show/Hide Solution
Finding the general solution:
The auxiliary equation is
[asciimath]r^2-8r+15=0[/asciimath]
The equation is factorable to
[asciimath](r-5)(r-3)=0[/asciimath]
The roots are [asciimath]r_1=5[/asciimath] and [asciimath]r_2=3[/asciimath] . This is Case 1 since the roots are real and distinct. Therefore, the general solution is the linear combination of [asciimath]y_1=e^(5x)[/asciimath] and [asciimath]y_2=e^(3x)[/asciimath]:
[asciimath]y(x)=c_1e^(5x)+c_2e^(3x)[/asciimath]
Applying the initial conditions:
Applying the initial condition to [asciimath]y[/asciimath]:
[asciimath]y(0)=9[/asciimath]
[asciimath]c_1e^(0)+c_2e^(0)=9[/asciimath]
[asciimath]c_1+c_2=9[/asciimath]
Applying the initial condition to [asciimath]y'[/asciimath]:
[asciimath]y'=5c_1e^(5x)+3c_2e^(3x)[/asciimath]
[asciimath]y'(0) = 35[/asciimath]
[asciimath]5c_1e^(0)+3c_2e^(0)=35[/asciimath]
[asciimath]5c_1+3c_2=35[/asciimath]
To determine [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath], we solve the following system of two equations and two unknowns:
[asciimath]{(c_1+c_2=9),(5c_1+3c_2=35 ):}[/asciimath]
Solving the system yields
[asciimath]c_1=4,[/asciimath] [asciimath]c_2=5[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]y(x)=4e^(5x)+5e^(3x)[/asciimath]
Try an Example
Find the general solution to the differential equation
[asciimath]y'' - 8 y' +16y = 0[/asciimath]
Show/Hide Solution
The auxiliary equation is
[asciimath]r^2-8r+16=0[/asciimath]
The equation is factorable to
[asciimath](r-4)^2=0[/asciimath]
The equation has a repeated root [asciimath]r=4[/asciimath]. This is Case 2, the repeated root. Therefore, the general solution is the linear combination of [asciimath]e^(4x)[/asciimath] and [asciimath]xe^(4x)[/asciimath] :
[asciimath]y(x)=c_1e^(4x)+c_2xe^(4x)[/asciimath]
Solve the following initial value problem (IVP).
[asciimath]y'' +10 y' + 25 y = 0,[/asciimath] [asciimath]y(0) = 4, \quad y'(0) = -25[/asciimath]
Show/Hide Solution
Finding the general solution:
The auxiliary equation is
[asciimath]r^2+10r+25=0[/asciimath]
The equation is factorable to
[asciimath](r+5)^2=0[/asciimath]
The equation has a repeated root [asciimath]r=-5[/asciimath]. This is Case 2, the repeated root. Therefore, the general solution is the linear combination of [asciimath]e^(-5x)[/asciimath] and [asciimath]xe^(-5x)[/asciimath]:
[asciimath]y(x)=c_1e^(-5x)+c_2xe^(-5x)[/asciimath]
Applying the initial conditions:
Applying the initial condition to [asciimath]y[/asciimath]:
[asciimath]y(0)=4[/asciimath]
[asciimath]c_1e^(0)+c_2(0)e^(0)=4[/asciimath]
[asciimath]c_1=4[/asciimath]
Applying the initial condition to [asciimath]y'[/asciimath]:
[asciimath]y'=-5c_1e^(-5x)+c_2(e^(-5x)-5xe^(-5x))[/asciimath]
[asciimath]y'(0) = -25[/asciimath]
[asciimath]-5c_1e^(0)+c_2(e^(0)-0)=-25[/asciimath]
[asciimath]-5c_1+c_2=-25[/asciimath]
Plugging in [asciimath]c_1=4[/asciimath] yields [asciimath]c_2=-5[/asciimath].
Therefore the solution to the initial value problem is
[asciimath]y(x)=4e^(-5x)-5xe^(-5x)[/asciimath]
Try an Example
Find the general solution to the differential equation
[asciimath]y'' - 4y' + 13y = 0[/asciimath]
Show/Hide Solution
The auxiliary equation is
[asciimath]r^2-4r+13=0[/asciimath]
Using the quadratic formula, we obtain
[asciimath]r=(4+-sqrt(16-52))/2=[/asciimath] [asciimath](4+-sqrt(-36))/2=2+-3i[/asciimath]
The equation has complex conjugate roots with a real part [asciimath]alpha=2[/asciimath] and an imaginary part [asciimath]beta=3[/asciimath]. This is Case 3 and thus the general solution is
[asciimath]y(x)=e^(2x)(c_1 cos(3x)+c_2 sin(3x))[/asciimath]
Solve the following initial value problem (IVP).
[asciimath]y'' +2 y' + 26 y = 0,[/asciimath] [asciimath]y(0) = -3, \quad y'(0) = -7[/asciimath]
Show/Hide Solution
Finding the general solution:
The auxiliary equation is
[asciimath]r^2+2r+26=0[/asciimath]
Alternative to using the quadratic formula that we used in the previous example, we can find the roots by completing the square. For variety, we use completing the square this time.
[asciimath]r^2+2r+1+25=0[/asciimath]
[asciimath]r^2+2r+1=-25[/asciimath]
[asciimath](r+1)^2=-25[/asciimath]
[asciimath]r+1=+-5i[/asciimath]
[asciimath]r=-1+-5i[/asciimath]
The equation has complex conjugate roots with a real part [asciimath]alpha=-1[/asciimath] and an imaginary part [asciimath]beta=5[/asciimath]. This is Case 3 and thus the general solution is
[asciimath]y(x)=e^(-x)(c_1 cos(5x)+c_2 sin(5x))[/asciimath]
Applying the initial conditions:
Applying the initial condition to [asciimath]y[/asciimath]:
[asciimath]y(0)=-3[/asciimath]
[asciimath]e^(0)(c_1 cos(0)+c_2 sin(0))=-3[/asciimath]
[asciimath]c_1=-3[/asciimath]
Applying the initial condition to [asciimath]y'[/asciimath]:
[asciimath]y'(x)=-e^(-x)(c_1 cos(5x)+c_2 sin(5x))+[/asciimath] [asciimath]e^(-x)(-5c_1 sin(5x)+5c_2 cos(5x))[/asciimath]
[asciimath]y'(0) = -7[/asciimath]
[asciimath]-c_1+5c_2=-7[/asciimath]
Plugging in [asciimath]c_1=-3[/asciimath] yields [asciimath]c_2=-2[/asciimath].
Therefore the solution to the initial value problem is
[asciimath]y(x)=e^(-x)(-3 cos(5x)-2 sin(5x))[/asciimath]
Try an Example
Section 3.2 Exercises
- Solve the given initial value problem.
[asciimath]2y''+11y'+12 y = 0,[/asciimath] [asciimath]y(0) = 4, \quad y'(0) = -8.5[/asciimath]
Show/Hide Answer
[asciimath]y(x)=e^(-4x)+3 e^(-1.5x)[/asciimath]
- Solve the given initial value problem.
[asciimath]y'' +4 y' + 4 y = 0,[/asciimath] [asciimath]y(0) = -1, \quad y'(0) = 8[/asciimath]
Show/Hide Answer
[asciimath]y(x)=-e^(-2x)+6xe^(-2x)[/asciimath]
- Solve the given initial value problem.
[asciimath]y'' +10 y' + 26 y = 0,[/asciimath] [asciimath]y(0) = 4, \quad y'(0) = -18[/asciimath]
Show/Hide Answer
[asciimath]y(x)=e^(-5x)(4cos(x)+2sin(x))[/asciimath]
