Differential Equations Cheatsheet 2nd-order Homogeneous
F (y 00 , y 0 , y, x) = 0 y 00 + a(x)y 0 + b(x)y = 0 Sol: yh = c1 y1 (x) + c2 y2 (x)
Jargon
General Solution: a family of functions, has parameters. Constant Coefficients
Reduction of Order - Method
Particular Solution: has no arbitrary parameters.
A.E. λ2 + aλ + b = 0
Singular Solution: cannot be obtained from the general solution. If we already know y1 , put y2 = vy1 ,
expand in terms of v 00 , v 0 , v, and put z = v 0 A. Real roots
and solve the reduced equation. Sol: y(x) = C1 eλ1 x + C2 eλ2 x
Linear Equations
Wronskian (Linear Independence) B. Single root
y (n) (x) + an−1 (x)y (n−1) (x) + · · · + a1 (x)y 0 (x) + a0 (x)y(x) = f (x) Sol: y(x) = C1 eλx + C2 xeλx
y1 (x) and y2 (x) are linearly independent iff
C. Complex roots
1st-order y1 y2
R R W (y1 , y2 )(x) = 6= 0 Sol: y(x) = eαx (C1 cos√βx + C2 sin βx)
F (y 0 , y, x) = 0 y 0 + a(x)y = f (x) I.F. = e a(x)dx
Sol: y = Ce− a(x)dx y10 y20 2
with α = − a2 and β = 4b−a2
Euler-Cauchy Equation A. Real roots
Variable Separable Homogeneous of degree 0
x2 y 00 + axy 0 + by = 0 where x 6= 0 Sol: y(x) = C1 xλ1 + C2 xλ2 x 6= 0
dy f (tx, ty) = t0 f (x, y) = f (x, y)
= f (x, y) A(x)dx + B(y)dy = 0 B. Single root
dx A.E. : λ(λ − 1) + aλ + b = 0
Test: Sol: Reduce to var.sep. using: Sol: y(x) = xλ (C1 + C2 ln |x|)
Sol: y(x) of the form xλ
f (x, y)fxy (x, y) = fx (x, y)fy (x, y) dy dv Reduction to Constant Coefficients: Use x = et , t = ln x, C. Complex roots (λ1,2 = α ± iβ)
y = xv =v+x
Sol: Separate and integrate on both sides. dx dx and rewrite in terms of t using the chain rule. Sol: y(x) = xα [C1 cos(β ln |x|) + C2 sin(β ln |x|)]
Exact Bernoulli
2nd-order Non-Homogeneous
M (x, y)dx + N (x, y)dy = 0 = dg(x, y) y 0 + p(x)y = q(x)y n
1 1 F (y 00 , y 0 , y, x) = 0 y 00 + a(x)y 0 + b(x)y = f (x) Sol: y = yh + yp = C1 y1 (x) + C2 y2 (x) + yp (x)
∂M ∂N Sol: Change var z = and divide by .
Iff = y n−1 yn
∂y ∂x
Sol: Find g(x, y) by integrating and comparing: Reduction by Translation Simple case: y 0 , y missing Simple case: y missing
Z Z
Ax + By + C 00
y = f (x) y 00 = f (y 0 , x)
M dx and N dy y0 =
Dx + Ey + F
Sol: Integrate twice. Sol: Change of var: p = y 0 and then solve twice.
Case I: Lines intersect
Reduction to Exact via Integrating Factor
Sol: Put x = X + h and y = Y + k, Simple case: y 0 , x missing Simple case: x missing
I(x, y)[M (x, y)dx + N (x, y)dy] = 0 find h and k, solve var.sep. and translate back.
y 00 = f (y) y 00 = f (y 0 , y)
Case I Case II: Parallel Lines (A = B, D = E)
Sol: Change of var: p = y 0 + chain rule, then Sol: Change of var: p = y 0 + chain rule, then
M y − Nx R u0 − A dp dp
If ≡ h(y) then I(x, y) = e− h(y)dx
Sol: Put u = Ax + By, y 0 = and solve. p = f (y) is var.sep. p = f (p, y) is 1st-order ODE.
M B dy dy
Case II Solve it, back-replace p and solve again. Solve it, back-replace p and solve again.
Nx − My R
If ≡ g(x) then I(x, y) = e− g(x)dx
Method of Undetermined Coefficients / “Guesswork”
N Variation of Parameters (Lagrange Method)
Case III Sol: Assume y(x) has same form as f (x) with
(More general, but you need to know yh )
undetermined constant coefficients.
If M = yf (xy) and N = xg(xy) then I(x, y) = Sol: yp = v1 y1 + v2 y2 + · · · + vn yn
1 Valid forms: v10 y1 + ··· + vn0 yn = 0
xM −yN
1. Pn (x) v20 y20 + ··· + vn0 yn0 = 0
··· + ··· + ··· = 0
2. Pn (x)eax (n−1) (n−1)
vn0 yb + ··· + vn0 yn = φ(x)
3. eax (Pn (x) cos bx + Qn (x)sinbx
Principle of Superposition
Failure case: If any term of f (x) is a solution of yh , Solve for all vi0 and integrate.
y 00 +ay 0 +by = f1 (x) has solution y1 (x) y 00 + ay 0 + by = f (x) = f1 (x) + f2 (x)
If then multiply yp by x and repeat until it works.
y 00 +ay 0 +by = f2 (x) has solution y2 (x) has solution: y(x) = y1 (x) + y2 (x)
Power Series Solutions
P∞
1. Assume y(x) = n=0 cn (x − a)n , compute y’, y” Taylor Series variant
2. Replace in the original D.E.
1. Differentiate both sides of the D.E. repeatedly
3. Isolate terms of equal powers
2. Apply initial conditions
4. Find recurrence relationship between the coefs.
3. Substitute into T.S.E. for y(x)
5. Simplify using common series expansions
(Use y = vx, z = v 0 to find y2 (x) if only y1 (x) is known.)
Validity
Regular singular point:
For y 00 + a(x)y 0 + b(x)y = 0
if xa(x) and x2 b(x) have a convergent MacLaurin
if a(x) and b(x) are analytic in |x| < R,
series near point x = 0. (Use translation if neces-
the power series also converges in |x| < R.
sary.)
Ordinary Point: Power method success guaranteed.
Irregular singular point: otherwise.
Singular Point: success not guaranteed.
Method of Frobenius for Regular Singular pt.
Case II: r1 = r2
∞
X P∞
y1 (x) = |x|r ( n=0 cn xn ) , c0 = 1
y(x) = xr (c0 + c1 x + c2 x2 + · · · ) = cn xr+n P ∞
n=0 y2 (x) = |x|r ( n=1 c∗n xn ) + y1 (x)ln|x|
Indicial eqn: r(r − 1) + a0 r + b0 = 0 Case III: r1 and r2 differ by an integer
P∞
Case I: r1 and r2 differ but not by an integer y1 (x) = |x|r1 ( n=0 cn xn ) , c0 = 1
P∞ P∞ ∗ n
y1 (x) = |x|r1 ( n=0 cn xn ) , c0 = 1 y2 (x) = |x| ( n=0 cn x ) + c∗1 y1 (x)ln|x|,
r2
c∗0 = 1
P∞
y2 (x) = |x|r2 ( n=0 c∗n xn ) , 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.