Learning Objectives
- Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.
Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form
ay′′+by′+cy=0ay′′+by′+cy=0
where aa, bb, and cc are constants.
Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let’s see what happens when we try a solution of the form y(x)=eλxy(x)=eλx, where λλ (the lowercase Greek letter lambda) is some constant.
If y(x)=eλxy(x)=eλx, then y′(x)=λeλxy′(x)=λeλx and y′′(x)=λ2eλxy′′(x)=λ2eλx. Substituting these expressions into a linear second-order differential equation, we get
ay′′+by′+cy=a(λ2eλx)+b(λeλx)+ceλx=eλx(aλ2+bλ+c)ay′′+by′+cy=a(λ2eλx)+b(λeλx)+ceλx=eλx(aλ2+bλ+c).
Since eλxeλx is never zero, this expression can be equal to zero for all xx only if
aλ1+bλ+c=0aλ1+bλ+c=0.
We call this the characteristic equation of the differential equation.
definition
The characteristic equation of the differential equation ay′′+by′+cy=0ay′′+by′+cy=0 is aλ2+bλ+c+0aλ2+bλ+c+0.
The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula
λ=−b±√b2−4ac2aλ=−b±√b2−4ac2a.
This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.
Distinct Real Roots
If the characteristic equation has distinct real roots λ1λ1 and λ2λ2, then eλ1xeλ1x and eλ2xeλ2x are linearly independent solutions to Example “Classifying Second-Order Equations”, and the general solution is given by
y(x)=c1eλ1x+c2eλ2xy(x)=c1eλ1x+c2eλ2x,
where c1c1 and c2c2 are constants.
For example, the differential equation y′′9y′+14y=0y′′9y′+14y=0 has the associated characteristic equation λ2+9λ+14=0λ2+9λ+14=0. This factors into (λ+2)(λ+7)=0(λ+2)(λ+7)=0, which has roots λ1=−2λ1=−2 and λ2=−7λ2=−7. Therefore, the general solution to this differential equation is
y(x)=c1e−2+c2e−7xy(x)=c1e−2+c2e−7x.
Single Repeated Real Root
Things are a little more complicated if the characteristic equation has a repeated real root, λλ. In this case, we know eλxeλx is a solution to Example “Classifying Second-Order Equations”, but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form keλxkeλx, where kk is some constant, but it would not be linearly independent of eλxeλx. Therefore, let’s try xeλxxeλx as the second solution. First, note that by the quadratic formula,
λ=−b±√b2−4ac2aλ=−b±√b2−4ac2a.
But, λλ is a repeated root, so b2−4ac=0b2−4ac=0 and λ=−b2aλ=−b2a. Thus, if y=xeλxy=xeλx, we have
y′=eλx+λxeλx and y′′=2λeλx+λ2xeλxy′=eλx+λxeλx and y′′=2λeλx+λ2xeλx.
Substituting these expressions into a linear second-order differential equation, we see that
ay′′+by′+cy=a(2λeλx+λ2eλx)+b(eλx+λxeλx)+cxeλx=xeλx(aλ2+bλ+c)+eλx(2aλ+b)=xeλx(0)+eλx(2(−b2a)+b)=0+eλx(0)=0ay′′+by′+cy=a(2λeλx+λ2eλx)+b(eλx+λxeλx)+cxeλx=xeλx(aλ2+bλ+c)+eλx(2aλ+b)=xeλx(0)+eλx(2(−b2a)+b)=0+eλx(0)=0.
This shows that xeλxxeλx is a solution to a linear second-order differential equation. Since eλxeλx and xeλxxeλx are linearly independent, when the characteristic equation has a repeated root λλ, the general solution to a linear second-order differential equation is given by
y(x)=c1eλx+c2xeλxy(x)=c1eλx+c2xeλx,
where c1c1 and c2c2 are constants.
For example, the differential equation y′′+12y′+36y=0y′′+12y′+36y=0 has the associated characteristic equation λ2+12λ+36=0λ2+12λ+36=0. This factors into (λ+6)2(λ+6)2, which has a repeated root λ=−6λ=−6. Therefore, the general solution to this differential equation is
Complex Conjugate Roots
The third case we must consider is when b2−4ac<0b2−4ac<0. In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number i=√−1i=√−1 to find the roots, which take the form λ1=α+βiλ1=α+βi and λ2=α−βiλ2=α−βi. The complex number α+βiα+βi is called the conjugate of α−βiα−βi. Thus, we see that when b2−4ac<0b2−4ac<0, the roots of our characteristic equation are always complex conjugates.
This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots—using the roots of the characteristic equation as the coefficients in the exponents of exponential functions—we get the functions e(α+βi)xe(α+βi)x and e(α−βi)xe(α−βi)x as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we’d like to find two linearly independent real-value solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions e(α+βi)xe(α+βi)x and e(α−βi)xe(α−βi)x as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.
Based on the roots α±βiα±βi of the characteristic equation, the functions e(α+βi)xe(α+βi)x and e(α−βi)xe(α−βi)x are linearly independent solutions to the differential equation. and the general solution is given by
y(x)=c1e(α+βi)x+c2e(α−βi)xy(x)=c1e(α+βi)x+c2e(α−βi)x.
Using some smart choices for c1c1 and c2c2, and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to a linear second-order differential equation and express our general solution in those terms.
We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was Euler’s formula, which tells us that
eiθ=cosθ+isinθeiθ=cosθ+isinθ
for all real numbers θθ.
Going back to the general solution, we have
y(x)=c1e(α+βi)x+c2e(α−βi)x=c1eαxeβix+c2eαxe−βix=eαx(c1eβix+c2e−βix)y(x)=c1e(α+βi)x+c2e(α−βi)x=c1eαxeβix+c2eαxe−βix=eαx(c1eβix+c2e−βix).
Applying Euler’s formula together with the identities cos(−x)=cosxcos(−x)=cosx and sin(−x)=−sinxsin(−x)=−sinx, we get
y(x)=eαx[c1(cosβx+isinβx)+c2(cos(−βx)+isin(−βx))]=eαx[(c1+c2)cosβx+(c1−c2)isinβx]y(x)=eαx[c1(cosβx+isinβx)+c2(cos(−βx)+isin(−βx))]=eαx[(c1+c2)cosβx+(c1−c2)isinβx].
Now, if we choose c1=c2=12, the second term is zero and we get
y(x)=eαxcosβx
as a real-value solution to a linear second-order differential equation. Similarly, if we choose c1=−i2 and c2=i2, the first term is zero and we get
y(x)=eαxsinβx
as a second, linearly independent, real-value solution to a linear second-order differential equation.
Based on this, we see that if the characteristic equation has complex conjugate roots α±βi, then the general solution to a linear second-order differential equation is given by
y(x)=c1eαxcosβx+c2eαxcossinβx=eαx(c1cosβx+c2sinβx).
where c1 and c2 are constants.
For example, the differential equation y′′−2y′+5y=0 has the associated characteristic equation λ2−2λ+5=0. By the quadratic formula, the roots of the characteristic equation are 1±2i. Therefore, the general solution to this differential equation is
y(x)=ex(c1cos2x+c2sin2x).
Summary of Results
We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in Table 7.1 Summary of Characteristic Equation Cases below.
Characteristic Equation Roots | General Solution to the Differential Equation |
---|---|
Distinct real roots, λ1 and λ2 | y(x)=c1eλ1x+c2eλ2x |
A repeated real root, λ | y(x)=c1eλx+c2xeλx |
Complex conjugate roots α±βi | y(x)=eλx(c1cosβx+c2sinβx) |
problem-solving strategy using the characteristic equation to solve second order differential equations with constant coefficients
- Write the differential equation in the form ay″+by′+cy=0.
- Find the corresponding characteristic equation aλ2+bλ+c=0.
- Either factor the characteristic equation or use the quadratic formula to find the roots.
- Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.
Example: solving second-order equations with constant coefficients
Find the general solution to the following differential equations. Give your answers as functions of x.
- y″+3y′−4y=0
- y″+6y′+13y=0
- y″+2y′+y=0
- y″−5y′=0
- y″−16y=0
- y″+16y=0
try it
Find the general solution to the following differential equations:
- y″−2y′+10y=0
- y″+14y′+49y=0
Watch the following video to see the worked solution to the above Try It
Candela Citations
- CP 7.6. Authored by: Ryan Melton. License: CC BY: Attribution
- Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/books/calculus-volume-3/pages/1-introduction. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction