Differential Equations Cheatsheet, Cheat Sheet of Mathematics

Download Differential Equations Cheatsheet and more Cheat Sheet Mathematics in PDF only on Docsity! Differential Equations Cheatsheet Jargon General Solution: a family of functions, has parameters. Particular Solution: has no arbitrary parameters. Singular Solution: cannot be obtained from the general solution. Linear Equations y(n)(x) + an−1(x)y(n−1)(x) + · · ·+ a1(x)y′(x) + a0(x)y(x) = f(x) 1st-order F (y′, y, x) = 0 y′ + a(x)y = f(x) I.F. = e R a(x)dx Sol: y = Ce− R a(x)dx Variable Separable dy dx = f(x, y) A(x)dx+B(y)dy = 0 Test: f(x, y)fxy(x, y) = fx(x, y)fy(x, y) Sol: Separate and integrate on both sides. Exact M(x, y)dx+N(x, y)dy = 0 = dg(x, y) Iff ∂M ∂y = ∂N ∂x Sol: Find g(x, y) by integrating and comparing:∫ Mdx and ∫ Ndy Reduction to Exact via Integrating Factor I(x, y)[M(x, y)dx+N(x, y)dy] = 0 Case I If My −Nx M ≡ h(y) then I(x, y) = e− R h(y)dx Case II If Nx −My N ≡ g(x) then I(x, y) = e− R g(x)dx Case III If M = yf(xy) and N = xg(xy) then I(x, y) = 1 xM−yN Homogeneous of degree 0 f(tx, ty) = t0f(x, y) = f(x, y) Sol: Reduce to var.sep. using: y = xv dy dx = v + x dv dx Bernoulli y′ + p(x)y = q(x)yn Sol: Change var z = 1 yn−1 and divide by 1 yn . Reduction by Translation y′ = Ax+By + C 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. Case II: Parallel Lines (A = B,D = E) Sol: Put u = Ax+By, y′ = u′ −A B and solve. Principle of Superposition If y ′′+ay′+by = f1(x) has solution y1(x) y′′+ay′+by = f2(x) has solution y2(x) then y ′′+ ay′+ by = f(x) = f1(x) + f2(x) has solution: y(x) = y1(x) + y2(x) 2nd-order Homogeneous F (y′′, y′, y, x) = 0 y′′ + a(x)y′ + b(x)y = 0 Sol: yh = c1y1(x) + c2y2(x) Reduction of Order - Method If we already know y1, put y2 = vy1, expand in terms of v′′, v′, v, and put z = v′ and solve the reduced equation. Wronskian (Linear Independence) y1(x) and y2(x) are linearly independent iff W (y1, y2)(x) = ∣∣∣∣ y1 y2y′1 y′2 ∣∣∣∣ 6= 0 Constant Coefficients A.E. λ2 + aλ+ b = 0 A. Real roots Sol: y(x) = C1eλ1x + C2eλ2x B. Single root Sol: y(x) = C1eλx + C2xeλx C. Complex roots Sol: y(x) = eαx(C1 cosβx+ C2 sinβx) with α = −a2 and β = √ 4b−a2 2 Euler-Cauchy Equation x2y′′ + axy′ + by = 0 where x 6= 0 A.E. : λ(λ− 1) + aλ+ b = 0 Sol: y(x) of the form xλ Reduction to Constant Coefficients: Use x = et, t = lnx, and rewrite in terms of t using the chain rule. A. Real roots Sol: y(x) = C1xλ1 + C2xλ2 x 6= 0 B. Single root Sol: y(x) = xλ(C1 + C2 ln |x|) C. Complex roots (λ1,2 = α± iβ) Sol: y(x) = xα [C1 cos(β ln |x|) + C2 sin(β ln |x|)] 2nd-order Non-Homogeneous F (y′′, y′, y, x) = 0 y′′ + a(x)y′ + b(x)y = f(x) Sol: y = yh + yp = C1y1(x) + C2y2(x) + yp(x) Simple case: y′, y missing y′′ = f(x) Sol: Integrate twice. Simple case: y′, x missing y′′ = f(y) Sol: Change of var: p = y′ + chain rule, then p dp dy = f(y) is var.sep. Solve it, back-replace p and solve again. Simple case: y missing y′′ = f(y′, x) Sol: Change of var: p = y′ and then solve twice. Simple case: x missing y′′ = f(y′, y) Sol: Change of var: p = y′ + chain rule, then p dp dy = f(p, y) is 1st-order ODE. 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. Variation of Parameters (Lagrange Method) (More general, but you need to know yh) Sol: yp = v1y1 + v2y2 + · · ·+ vnyn v′1y1 + · · · + v′nyn = 0 v′2y ′ 2 + · · · + v′ny′n = 0 · · · + · · · + · · · = 0 v′ny (n−1) b + · · · + v′ny(n−1)n = φ(x) Solve for all v′i and integrate.