Methods of Solution of Selected Differential Equations

Carol A. Edwards
Chandler-Gilbert Community College

Equations of Order One: Mdx + Ndy = 0

1. Separate variables.

2. M, N homogeneous of same degree:

Substitute y = vx or x = vy
dy = vdx + xdv dx = vdy + ydv
and then separate variables.

3. Exact: ∂M = ∂N
∂y ∂x
Solve ∂F = M for F(x,y) including f(y) as constant term.
∂x
Then compute ∂F = N to find f(y).
∂y
Solution is F(x,y) = c.
Alternatively, start with ∂F = N.
∂y

4. Linear: dy + P(x)y = Q(x)

dx [IF=Integrating Factor]
IF = exp( ∫ Pdx)
Multiply both sides of the equation by IF and result is exact.
Left hand side will be d (IF•y)
dx

5. The orthogonal trajectories to the family that has differential equation

Mdx + Ndy = 0 have differential equation Ndx - Mdy = 0.

6. IF by inspection:
y
Look for d(xy) = xdy + ydx d( ) = xdy - ydx
x
x2
x -1 y
d( ) = ydx - xdy d(tan ) = xdy - ydx
y x
y2 x2 + y2
It may help to group terms of like degree.

7. IF for certain equations that are not homogeneous, not exact, and not linear:
a. If 1 ∂M - ∂N = f(x), a function of x alone.
N ∂y ∂x
IF = exp( ∫ f(x) dx). Resulting equation is exact.
b. If 1 ∂M - ∂N = g(y), a function of y alone.
M ∂y ∂x
IF = exp(- ∫ g(y) dy). Resulting equation is exact.
8. Substitution suggested by the equation:

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 If an expression appears more than once, substituting a single variable for it may reduce
the equation to a recognizable form.

9. Bernoulli: y′ + P(x)y = Q(x)yn

Substitute z = y1-n and the resulting equation will be linear in z.

10. Coefficients both linear:

(a1x + b1y + c1)dx + (a2x + b2y + c2)dy = 0
Consider lines a1x + b1y + c1 = 0
a2x + b2y + c2 = 0

a. If lines intersect at (h,k), substitute x = u + h, y = v + k

to get (a1u + b1v)du + (a2u + b2v)dv = 0 which is homogeneous.

b. If lines are parallel or coincide, use a substitution for recurring expression. (See 8)

Linear Differential Equation:

b0(x)dny + b1(x)dn-1y + . . . + bn-1(x)dy + bn(x)y = R(x)
dxn dxn-1 dx

1. The functions f1, f2, . . . , fn are linearly independent when

c1f1(x) + c2f2(x) + . . . + cnfn(x) = 0 implies c1 = c2 = . . . = cn = 0.

2. The functions f1, f2, . . . , fn are linearly dependent if there exist

constants c1, c2, . . . , cn, not all zero, such that
c1f1(x) + c2f2(x) + . . . + cnfn(x) = 0 identically on a ≤ x ≤ b.

3. The Wronskian of f1, f2, . . . , fn is f1 f2 f3 . . . fn

f1′ f2′ f3′ . . . fn′
f1′′ f2′′ f3′′ . . . fn′′
f1(n-1) f2(n-1) f3(n-1) . . . fn(n-1)

4. Theorem: If on (a,b), b0(x) ≠ 0, b1, b2, . . . , bn continuous,

and y1, y2, . . . , yn are solutions of
b0y(n) + b1y(n-1) + . . . + bn-1y′ + bny = 0
then y1, y2, . . . , yn are linearly independent if and only if the
Wronskian of y1, y2, . . . , yn is not zero on (a,b).

5. If y1, y2, . . . , yn are linearly independent solutions of the homogeneous equation,

b0y(n) + b1y(n-1) + . . . + bn-1y′ + bny = 0
then the general solution of this equation is y = c1y1 + c2y2 + . . . + cnyn.

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 6. The general solution of the equation
b0y(n) + b1y(n-1) + . . . + bn-1y′ + bny = R(x)
is y = yc + yp, where yc = c1y1 + c2y2 + . . . + cnyn,
the complementary function; y1, y2, . . . , yn are linearly independent solutions of the
homogeneous equation; and c1, c2, . . . , cn are arbitrary constants; and yp is any
particular solution of the given nonhomogeneous equation.

7. A differential operator of order n

A = a0Dn + a1Dn-1 + . . . + an-1D + an where Dky = dky
dxk

8. Properties of differential operators:

a. If f(D) is a polynomial in D, then f(D) [emx] = emxf(m).

b. If f(D) is a polynomial in D with constant coefficients,

eaxf(D)y = f(D-a) [eaxy] (“exponential shift”)

c. (D – m)n(xkemx) = 0 for k = 0, 1, . . . , (n-1).

Linear Equations with Constant Coefficients:

a0y(n) + a1y(n-1) + . . . + an-1y′ + any = R(x)
i.e., f(D)y = R(x)

1. The auxiliary equation associated with f(D)y = 0 is f(m) = 0.

a. f(m) = 0 has distinct real roots m1, m2, . . . , mn:
yc = c1em1 x + c2em2 x + . . . + cnemn x

b. f(m) = 0 has repeated real roots. For each set of repetitions,

k
say, b, b, . . . , b, the solutions are
c1ebx, c2xebx, c3x2ebx, . . . , ckxk-1ebx

c. f(m) = 0 has distinct imaginary roots:

For m = a ± bi, y = c1eaxcosbx + c2eaxsinbx

d. f(m) = 0 has repeated imaginary roots. For example for

a ± bi, a ± bi, y = (c1 + c2x)eaxcosbx + (c3 + c4x)eaxsinbx.

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 2. Method of undetermined coefficients:

a. m1, m2, . . . , mn solutions of the auxiliary equation, so

yc = c1y1 + . . . + cnyn

b. Assuming R(x) is itself a particular solution of some homogeneous differential

equation with constant coefficients which has roots m1’, m2’, . . . , mk’ for its
auxiliary equation. Write yp from m1’, m2’, . . . , mk’ being careful about any
repetitions of m’-values with m-values. Substitute this yp in the original equation,
f(D)y = R(x) and equate corresponding coefficients.

c. General solution: y = yc + yp.

3. Solutions by inspection:
a. If R(x) = constant and an ≠ 0 then yp = R(x)
an
b. If R(x) = constant and an = 0 with y(k) the lowest-order derivative that actually
appears, then yp = R(x)•xk
k! an-k
4. If y1 is a particular solution of f(D)y = R1(x) and y2 is a particular solution of
f(D)y = R2(x), then yp = y1 + y2 is a particular solution of f(D)y = R1(x) + R2(x).

Linear Equations with Variable or Constant Coefficients

(b0Dn + b1Dn-1 + . . . + bn-1D + bn)y = R(x), bi is not necessarily constant.

1. Reduction of order (d’Alembert): y′′ + py′ + qy = R

If y = y1 is a solution of the corresponding homogeneous equation:
y′′ + py′ + qy = 0.
Let y = vy1, v variable, and substitute into original equation and simplify.
Set v′ = w and the resulting equation is a linear equation of first order in w. Find the
IF and solve for w. Then since v’ = w, find v by integration. This gives y = vy1.

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 2. Variation of parameters (Lagrange)

a. Order two: y′′ + py′ + qy = R(x)

If yc = c1y1 + c2y2, set yp = A(x)y1 + B(x)y2, then find A and B so that
this is a particular solution of the nonhomogeneous equation.
A′y1 + B′y2 = 0
A′y1’ + B′y2’ = R(x)

Solve the system for A′ and B′, then for A and B by integration.
Then yp = A(x)y1 + B(x)y2.

b. Order three: y′′′ + py′′ + qy′ + r = s(x)

If yc = c1y1 + c2y2 + c3y3 then set yp = A(x)y1 + B(x)y2 + C(x)y3.
A′y1 + B′y2 + C′y3 = 0
A′y1′ + B′y2′ + C′y3′ = 0
A′y1′′ + B′y2′′ + C′y3′′ = s(x)
Solve the system for A′, B′, and C′, then for A, B, and C by integration.
Then yp = A(x)y1 + B(x)y2 + C(x)y3.

Inverse Differential Operators

1. Exponential shift: eaxf(D)y = f(D – a) [eaxy]

2. Evaluation of 1 eax
f(D)
a. If f(a) ≠ 0 then 1 eax = eax
f(D) f(a)
b. If f(a) = 0 then 1 eax = xneax , φ(a) ≠ 0.
n
φ(D)(D-a) n! φ(a)

3. Evaluation of (D2 + a2)-1sin ax and (D2 + a2)-1cos ax.

a. If a ≠ b, 1 sin bx = sin bx
D2 + a2 a2 – b2

1 cos bx = cos bx
D2 + a2 a2 – b2

b. If a = b, 1 sin ax = -x cos ax
D + a2
2
2a

1 cos ax = x sin ax
D2 + a2 2a

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Laplace Transform
∞
1. Definition: Laplace transform of F(t) = L{F(t)} = ∫0 e-stF(t)dt = f(s)

2. L is a linear transformation: c1, c2 constants

L{c1F1 + c2F2} = c1L{F1} + c2L{F2}.

3. Transforms of elementary functions.

a. L{ekt} = 1 , s > k
s–k

b. L{sin kt} = k , s > 0

s 2 + k2

L{cos kt} = s , s > 0

s + k2
2

c. L{tn} = n! , s > 0, n positive integer.

sn+1

4. Definition: A function F(t) is sectionally continuous over [a,b] if [a,b] can be

divided into a finite number of sub-intervals [c,d] such that in each subinterval:
(1) F(t) is continuous on [c,d], and
(2) lim F(t) and lim F(t) exist.
t→c+ t→d-

5. Definition: The function F(t) is of exponential order as t→∞ if there exist constants
M, b, and a fixed t-value t0 such that |F(t)| < Mebt for t ≥ t0.
a. Note: a bounded function is of exponential order as t→∞
b. If there is a b such that lim [e-btF(t)] exists, then F(t) is of exponential order
t→∞
as t→∞.

6. Definition: A function of Class A is any function that is

(1) sectionally continuous over every finite interval in the range t ≥ 0, and
(2) of exponential order as t→∞.

7. Theorem: If F(t) is a function of Class A, then L{F(t)} exists.

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 8. Solution of initial value problems.
Theorem: If F(t), F′ (t), . . . , F(n-1)(t) are continuous for t ≥ 0 and of exponential
n-1
order as t→∞ and if F(n)(t) is of Class A, then L{F(n)(t)} = snL{F(t)} -∑ sn-1-
k (k) n=0
F (0).
In particular
n = 1: L{F′(t)} = sL{F(t)} – F(0).
n = 2: L{F′′(t)} = s2L{F(t)} – sF(0) – F′(0).
n = 3: L{F′′′(t)} = s3L{F(t)} – s2F(0) – sF′(0) – F′′(0).

Theorem: If F(t) is of exponential order as t→∞ and F(t) is continuous for t ≥ 0

except for a finite jump at t = t1, and if F′(t) is of Class A, then from
L{F(t)) = f(s}, it follows that L{F′(t)} = sf(s) – F(0) – exp(-st1)[F(t1+) – F(t1-)]

9. Derivatives of transforms.
Theorem: If F(t) is of Class A, then for every positive integer n,
dn f(s) = L{(-t)nF(t)} where f(s) = L{F(t)}.
dsn

10. Transform of a periodic function.

Theorem: If F(t) is periodic with period ω and F(t) has a Laplace transform
ω
then L{F(t))} = ∫0 e-st F(t) dt
1 – e-sω

11. Definition: If L{F(t)} = f(s) then F(t) is an inverse transform of f(s)

and F(t) = L-1{f(s)}.

12. L-1 is a linear transformation.

13. Theorem: L-1{f(s)} = e-atL-1{f(s – a)}.

Gamma Function
∞
1. Definition: Γ(x) = ∫0 e-ββx-1dβ , x > 0.

2. Theorem: For all x > 0, Γ(x + 1) + xΓ(x).

3. Theorem: Γ(n + 1) = n! if n is a positive integer.

4. Theorem: L{tx} = Γ(x + 1) , s > 0, x > -1.

sx+1

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