Differential Equations Cheatsheet
2nd-order Homogeneous
F (y 00 , y 0 , y, x) = 0
Jargon
General Solution: a family of functions, has parameters.
Particular Solution: has no arbitrary parameters.
Singular Solution: cannot be obtained from the general solution.
Reduction of Order - Method
If we already know y1 , put y2 = vy1 ,
expand in terms of v 00 , v 0 , v, and put z = v 0
and solve the reduced equation.
Linear Equations
Wronskian (Linear Independence)
y (n) (x) + an−1 (x)y (n−1) (x) + · · · + a1 (x)y 0 (x) + a0 (x)y(x) = f (x)
1st-order
F (y 0 , y, x) = 0
y 0 + a(x)y = f (x)
Variable Separable
dy
= f (x, y)
dx
Test:
I.F. = e
R
a(x)dx
Sol: y = Ce−
a(x)dx
Euler-Cauchy Equation
f (tx, ty) = t0 f (x, y) = f (x, y)
A(x)dx + B(y)dy = 0
Sol: Reduce to var.sep. using:
f (x, y)fxy (x, y) = fx (x, y)fy (x, y)
dy
dv
=v+x
dx
dx
y = xv
Sol: Separate and integrate on both sides.
Bernoulli
M (x, y)dx + N (x, y)dy = 0 = dg(x, y)
Iff
∂M
∂N
=
∂y
∂x
Case II
Nx − My
If
≡ g(x)
N
I(x, y) = e−
then
I(x, y) = e−
R
R
1
.
yn
Ax + By + C
y0 =
Dx + Ey + F
Case I: Lines intersect
Sol: Put x = X + h and y = Y + k,
find h and k, solve var.sep. and translate back.
I(x, y)[M (x, y)dx + N (x, y)dy] = 0
then
and divide by
Reduction by Translation
Reduction to Exact via Integrating Factor
Case I
M y − Nx
If
≡ h(y)
M
1
y n−1
h(y)dx
Case II: Parallel Lines (A = B, D = E)
u0 − A
Sol: Put u = Ax + By, y 0 =
and solve.
B
g(x)dx
Case III
If M = yf (xy) and N = xg(xy) then I(x, y) =
1
xM −yN
Principle of Superposition
y 00 +ay 0 +by = f1 (x) has solution y1 (x)
y 00 + ay 0 + by = f (x) = f1 (x) + f2 (x)
then
y 00 +ay 0 +by = f2 (x) has solution y2 (x)
has solution: y(x) = y1 (x) + y2 (x)
x2 y 00 + axy 0 + by = 0 where x 6= 0
Sol: yh = c1 y1 (x) + c2 y2 (x)
Constant Coefficients
λ2 + aλ + b = 0
A.E.
A. Real roots
Sol: y(x) = C1 eλ1 x + C2 eλ2 x
B. Single root
Sol: y(x) = C1 eλx + C2 xeλx
C. Complex roots
Sol: y(x) = eαx (C1 cos√βx + C2 sin βx)
2
with α = − a2 and β = 4b−a
2
A. Real roots
Sol: y(x) = C1 xλ1 + C2 xλ2
x 6= 0
B. Single root
A.E. : λ(λ − 1) + aλ + b = 0
Sol: y(x) = xλ (C1 + C2 ln |x|)
Sol: y(x) of the form xλ
Reduction to Constant Coefficients: Use x = et , t = ln x, C. Complex roots (λ1,2 = α ± iβ)
and rewrite in terms of t using the chain rule.
Sol: y(x) = xα [C1 cos(β ln |x|) + C2 sin(β ln |x|)]
2nd-order Non-Homogeneous
y 0 + p(x)y = q(x)y n
Sol: Change var z =
Sol: Find g(x, y) by integrating and comparing:
Z
Z
M dx
and
N dy
If
R
y1 (x) and y2 (x) are linearly independent iff
y y2 6= 0
W (y1 , y2 )(x) = 10
y1 y20 Homogeneous of degree 0
Exact
y 00 + a(x)y 0 + b(x)y = 0
F (y 00 , y 0 , y, x) = 0
y 00 + a(x)y 0 + b(x)y = f (x)
Simple case: y 0 , y missing
Sol: y = yh + yp = C1 y1 (x) + C2 y2 (x) + yp (x)
Simple case: y missing
y 00 = f (y 0 , x)
00
y = f (x)
Sol: Change of var: p = y 0 and then solve twice.
Sol: Integrate twice.
Simple case: y 0 , x missing
Simple case: x missing
y 00 = f (y 0 , y)
y 00 = f (y)
Sol: Change of var: p = y 0 + chain rule, then
dp
p
= f (y) is var.sep.
dy
Solve it, back-replace p and solve again.
Method of Undetermined Coefficients / “Guesswork”
Sol: Assume y(x) has same form as f (x) with
undetermined constant coefficients.
Valid forms:
1. Pn (x)
2. Pn (x)eax
3. eax (Pn (x) cos bx + Qn (x)sinbx
Failure case: If any term of f (x) is a solution of yh ,
multiply yp by x and repeat until it works.
Sol: Change of var: p = y 0 + chain rule, then
dp
p
= f (p, y) is 1st-order ODE.
dy
Solve it, back-replace p and solve again.
Variation of Parameters (Lagrange Method)
(More general, but you need to know yh )
Sol: yp = v1 y1 + v2 y2 + · · · + vn yn
v10 y1
v20 y20
···
(n−1)
vn0 yb
+
+
+
+
···
···
···
···
+
+
+
+
vn0 yn
vn0 yn0
···
(n−1)
vn0 yn
Solve for all vi0 and integrate.
=
=
=
=
0
0
0
φ(x)
Power Series Solutions
1.
2.
3.
4.
5.
P∞
Assume y(x) = n=0 cn (x − a)n , compute y’, y”
Replace in the original D.E.
Isolate terms of equal powers
Find recurrence relationship between the coefs.
Simplify using common series expansions
Taylor Series variant
1. Differentiate both sides of the D.E. repeatedly
2. Apply initial conditions
3. Substitute into T.S.E. for y(x)
(Use y = vx, z = v 0 to find y2 (x) if only y1 (x) is known.)
Validity
For y 00 + a(x)y 0 + b(x)y = 0
if a(x) and b(x) are analytic in |x| < R,
the power series also converges in |x| < R.
Ordinary Point: Power method success guaranteed.
Singular Point: success not guaranteed.
Regular singular point:
if xa(x) and x2 b(x) have a convergent MacLaurin
series near point x = 0. (Use translation if necessary.)
Irregular singular point: otherwise.
Method of Frobenius for Regular Singular pt.
Case II: r1 = r2
y(x) = xr (c0 + c1 x + c2 x2 + · · · ) =
∞
X
cn xr+n
n=0
Indicial eqn:
r(r − 1) + a0 r + b0 = 0
Case I: r1 and r2 differ but not by an integer
P∞
y1 (x) = |x|r1 ( n=0 cn xn ) , c0 = 1
P∞
y2 (x) = |x|r2 ( n=0 c∗n xn ) , c∗0 = 1
y1 (x)
y2 (x)
P∞
= |x|r ( n=0 cn xn ) ,
P
∞
= |x|r ( n=1 c∗n xn ) + y1 (x)ln|x|
c0 = 1
Case III: r1 and r2 differ by an integer
P∞
y1 (x)
= |x|r1 ( n=0 cn xn ) ,
P∞ ∗ n
r2
y2 (x) = |x| ( n=0 cn x ) + c∗1 y1 (x)ln|x|,
c0 = 1
c∗0 = 1
Laplace Transform
FIXME TODO
Fourier Transform
FIXME TODO
Author: Martin Blais, 2009. This work is licensed under the Creative Commons “Attribution - Non-Commercial - Share-Alike” license.