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Higher Order Linear Differential Equations with Constant Coefficients
Part I. Homogeneous Equations: Characteristic Roots
Objectives: Solve n-th order homogeneous linear equations
any(n)^ + an− 1 y(n−1)^ + · · · + a 1 y′^ + a 0 y = 0, where an, · · · , a 1 , a 0 are constants with an 6 = 0.
Solution Method:
- Find the roots of the characteristic polynomial: anλn^ + an− 1 λn−^1 + · · · + a 1 λ + a 0 = 0.
- Each root λ produces a particular exponential solution eλt^ of the differential equation.
- A repeated root λ of multiplicity k produces k linearly independent solutions eλt, teλt, · · · , tk−^1 eλt.
Warning: The above method of characteristic roots does not work for linear equations with variable coefficients. As matter of fact, the explicit solution method does not exist for the general class of linear equations with variable coefficients.
Example 1: (a) Find general solutions of y′′′^ + 4y′′^ − 7 y′^ − 10 y = 0. (b) Solve y′′′^ + 4y′′^ − 7 y′^ − 10 y = 0, y(0) = − 3 , y′(0) = 12, y′′(0) = −36. Solution: (a) Solve the characteristic polynomial: λ^3 + 4λ^2 − 7 λ − 10 = 0. The roots are λ 1 = − 1 , λ 2 = 2, λ 3 = −5. Each root gives a particular exponential solution of the differential equation. Combined, the general solutions are y = C 1 e−t^ + C 2 e^2 t^ + C 3 e−^5 t, where C 1 , C 2 , C 3 are free parameters (arbitrary constants). (b) We use the initial conditions to determine the values of the constants C 1 , C 2 , C 3 in the general solution formula. The initial conditions y(0) = − 3 , y′(0) = 12, y′′(0) = −36 yield C 1 + C 2 + C 3 = − 3 , −C 1 + 2C 2 − 5 C 3 = 12, C 1 + 4C 2 + 25C 3 = − 36. Solve this linear system for C 1 , C 2 , C 3 : C 1 = − 5 / 2 , C 2 = 1, C 3 = − 3 / 2. Thus, the solution to the initial value problem is
y = −
e−t^ + e^2 t^ −
e−^5 t.
Example 2: (a) Find general solutions of y(4)^ + 8y′′^ + 16y = 0. (b) Solve y(4)^ + 8y′′^ + 16y = 0, y(0) = − 1 , y′(0) = 5, y′′(0) = − 8 , y′′′(0) = −28. Solution: (a) Solve the characteristic polynomial:
0 = λ^4 + 8λ^2 + 16 = (λ^2 + 4)^2.
The roots are λ 1 = 2i, λ 2 = 2i, λ 3 = − 2 i, λ 4 = − 2 i. We have repeated roots. For complex characteristic roots, we can either use complex exponential functions or use cos and sin to express the solutions. Expression 1: The general solutions are
y = C 1 cos(2t) + C 2 sin(2t) + C 3 t cos(2t) + C 4 t sin(2t),
where C 1 , C 2 , C 3 , C 4 are free parameters (arbitrary constants). Expression 2: The general solutions are
y = a 1 e^2 it^ + a 2 te^2 it^ + a 3 e−^2 it^ + a 4 te−^2 it,
where a 1 , a 2 , a 3 , a 4 are free parameters (arbitrary constants). (b) Let’s use Expression 1 in the above. The initial conditions yield a linear system for C 1 , · · · , C 4 : C 1 = − 1 , 2 C 2 +C 3 = 5 , − 4 C 1 +4C 4 = − 8 , − 8 C 2 − 12 C 3 = − 28. Solving this we obtain C 1 = − 1 , C 2 = 2, C 3 = 1, C 4 = −3. Thus, the solution is
y = − cos(2t) + 2 sin(2t) + t cos(2t) − 3 t sin(2t).
We’re done. Alternatively, we may use Expression 2 as well. The initial conditions yield
a 1 +a 3 = − 1 , 2 ia 1 +a 2 − 2 ia 3 +a 4 = 5 , − 4 a 1 +4ia 2 − 4 a 3 − 4 ia 4 = − 8 , − 8 ia 1 − 12 a 2 +8ia 3 − 12 a 4 = − 28.
Solving this we obtain a 1 = − 1 / 2 − i, a 2 = (1 + 3i)/ 2 , a 3 = − 1 /2 + i, a 4 = (1 − 3 i)/2. Thus, the solution is
y =
− i
e^2 it^ +
i
te^2 it^ +
e−^2 it^ +
i
te−^2 it.
Part II. Nonhomogeneous Equations: Undetermined Coefficients
Objectives: Solve n-th order nonhomogeneous linear equations
any(n)^ + an− 1 y(n−1)^ + · · · + a 1 y′^ + a 0 y = f (t),
where an, · · · , a 1 , a 0 are constants with an 6 = 0, and f (t) is a given function.
Solution Method:
- The general solutions of the nonhomogeneous equation are of the following structure:
y = yc + yp,
where yc (the so-called “complementary” solutions) are solutions of the corresponding homogeneous equation:
any( cn )+ an− 1 y c(n −1)+ · · · + a 1 y′ c + a 0 yc = 0,
and yp is a particular solution of the given nonhomogeneous equation.
- We copy the function structure of the nonhomogeneous term f (t) to set up a trial function for yp.
- A“naive” choice of the trial function yp does not work in the case of resonance; that is, we need to modify the choice of yp in the case when there are overlaps between yc and yp. The guiding principle of the modification is to multiply the corresponding term(s) by tm, where m is the smallest positive integer to make the overlap between yc and yp disappear.
Limitations: The method of undetermined coefficients does not work for linear equations with variable coefficients. Also the nonhomogeneous term f (t) is restricted to be of the following forms:
a polynomial p(t) = a 0 + a 1 t + · · · + aN tN^ , an exponential function eat, cos(ωt), sin(ωt), (a polynomial)·eat, (a polynomial)· cos(ωt), (a polynomial)· sin(ωt), (a polynomial)·eat^ cos(ωt), (a polynomial)·eat^ sin(ωt), or a linear combination of the above functions.
For those f (t) that are not one of the above, the method of variation of parameters should be used to solve the nonhomogeneous equation. Indeed, the method of variation of parameters is a more general method and works for arbitrary nonhomogeneous term f (t) (including the types that can be solved by the method of undetermined coefficients).
Example 4: (a) Solve y′′′^ + 4y′′^ − 7 y′^ − 10 y = 100t^2 − 64 e^3 t. (b) Solve y′′′^ + 4y′′^ − 7 y′^ − 10 y = 100t^2 − 64 e^3 t, y(0) = − 20 , y′(0) = − 29 / 5 , y′′(0) = 19/5. Solution: (a) First, solve the corresponding homogeneous equation y′′′ c + 4y′′ c − 7 y c′ − 10 yc = 0. This was done in Example 1. The complementary solutions are yc = C 1 e−t^ + C 2 e^2 t^ + C 3 e−^5 t, where C 1 , C 2 , C 3 are free parameters. Next, set up a trial function by copying the structure of f (t): yp = a 0 + a 1 t + a 2 t^2 + be^3 t. Substitute this into the nonhomogeneous equation and simplify: (− 10 a 0 − 7 a 1 + 8a 2 ) + (− 10 a 1 − 14 a 2 )t − 10 a 2 t^2 + 32be^3 t^ = 100t^2 − 64 e^3 t. Compare the coefficients of the two sides: − 10 a 0 − 7 a 1 + 8a 2 = 0, − 10 a 1 − 14 a 2 = 0, − 10 a 2 = 100, 32 b = − 64. Solving this linear system we obtain a 0 = − 89 / 5 , a 1 = 14, a 2 = − 10 , b = − 2 , and thus yp = −
- 14t − 10 t^2 − 2 e^3 t. The general solutions to the nonhomogeneous equation are
y = yp + yc = −
- 14t − 10 t^2 − 2 e^3 t^ + C 1 e−t^ + C 2 e^2 t^ + C 3 e−^5 t,
where C 1 , C 2 , C 3 are free parameters. (b) Examine the initial conditions:
C 1 + C 2 + C 3 −
−C 1 + 2C 2 − 5 C 3 + 14 − 6 = −
C 1 + 4C 2 + 25C 3 − 20 − 18 =
Solve this for C 1 , C 2 , C 3 : C 1 = −
, C 2 = − 2 , C 3 = 2.
Therefore, the solution of the initial value problem is
y = −
- 14t − 10 t^2 − 2 e^3 t^ −
e−t^ − 2 e^2 t^ + 2e−^5 t.
Example 6: Solve y(6)^ + y(5)^ − 3 y(4)^ − 5 y(3)^ − 2 y(2)^ = −4 + 8t^2 + 16tet^ + (54 − 18 t − 9 t^4 )e−t.
Solution: First, solve the corresponding homogeneous equation
y c(6) + y c(5) − 3 y(4) c − 5 y(3) c − 2 y(2) c = 0.
Find the roots of the characteristic polynomial:
λ^6 + λ^5 − 3 λ^4 − 5 λ^3 − 2 λ^2 = λ^2 (λ + 1)^3 (λ − 2).
Hence, the complementary solutions are
yc = C 1 + C 2 t + C 3 e−t^ + C 4 te−t^ + C 5 t^2 e−t^ + C 6 e^2 t,
where C 1 , · · · , C 6 are free parameters. Next, set up a trial form for a particular solution:
yp = t^2 (a 0 + a 1 t + a 2 t^2 ) + (b 0 + b 1 t)et^ + t^3 (c 0 + c 1 t + c 2 t^2 + c 3 t^3 + c 4 t^4 )e−t.
Notice that this is a resonant case and we needed the red terms to modify yp. Plugging yp in the nonhomogeneous equation and comparing the coefficients of the two sides, we can determine the coefficients: a 0 = − 18 , a 1 = 10/ 3 , a 2 = − 1 / 3 , b 0 = 5, b 1 = − 2 , c 0 = 2170/ 27 , c 1 = 577/ 36 , c 2 = 34/ 15 , c 3 = 7/ 30 , c 3 = 1/ 70. Therefore, the general solutions of the nonhomogeneous equation are
y = yp + yc
= t^2
t −
t^2
t +
t^2 +
t^3 +
t^4
e−t
+C 1 + C 2 t + C 3 e−t^ + C 4 te−t^ + C 5 t^2 e−t^ + C 6 e^2 t,
where C 1 , · · · , C 6 are free parameters.
EXERCISES
Solve the following differential equations and initial value problems.
[1] (a) y′′′^ + 4y′′^ − y′^ − 4 y = 0 (b) y′′′^ + 4y′′^ − 3 y′^ − 18 y = 0 (c) y′′′^ + 6y′′^ + 12y′^ + 8y = 0
[2] y′′′^ − 5 y′′^ − y′^ + 5y = 10t − 63 e−^2 t^ + 29 sin(2t).
[3] y(4)^ +2y′′′^ − 2 y′^ −y = 24te−t^ +24et^ −8 sin t, y(0) = − 2 , y′(0) = 0, y′′(0) = 6, y′′′(0) = 10.
[4] y(4)^ + 18y′′^ + 81y = 64t cos t + 108t cos(3t)
Answers:
[1] (a) y = C 1 e−t^ + C 2 et^ + C 3 e−^4 t^ (b) y = C 1 e^2 t^ + C 2 e−^3 t^ + C 3 te−^3 t (c) y = C 1 e−^2 t^ + C 2 te−^2 t^ + C 3 t^2 e−^2 t
[2] y =
cos(2t) + sin(2t) + C 1 e−t^ + C 2 et^ + C 3 e^5 t
[3] y = t^3
t
e−t^ + 3tet^ − 2 cos t − et^ + e−t^ − te−t^ − 2 t^2 e−t
[4] y = t cos t +
sin t −
t^3 cos(3t) +
t^2 sin(3t) + C 1 cos(3t) + C 2 sin(3t) + C 3 t cos(3t) + C 4 t sin(3t)
