MATH 202 Useful File Summary, Spring 2016

Differential Equations Study Guide1 First Order Equations General Form of ODE: dy dx = f(x, y) (1) Initial Value Problem: y′ = f(x, y), y(x0) = y0 (2) Linear Equations General Form: y′ + p(x)y = f(x) (3) Integrating Factor: µ(x) = e � p(x)dx (4) =⇒ d dx (µ(x)y) = µ(x)f(x) (5) General Solution: y = 1 µ(x) �� µ(x)f(x)dx + C � (6) Homogeneous Equations General Form: y′ = f(y/x) (7) Substitution: y = zx (8) =⇒ y′ = z + xz′ (9) The result is always separable in z: (10) dz f(z) − z = dx x Bernoulli Equations General Form: y′ + p(x)y = q(x)yn (11) Substitution: z = y1−n (12) The result is always linear in z: (13) z′ + (1 − n)p(x)z = (1 − n)q(x) Exact Equations General Form: M(x, y)dx + N(x, y)dy = 0 (14) Text for Exactness: ∂M ∂y = ∂N ∂x (15) Solution: φ = C where (16) M = ∂φ ∂x and N = ∂φ ∂y (17) Method for Solving Exact Equations: 1. Let φ = � M(x, y)dx + h(y) 2. Set ∂φ ∂y = N(x, y) 3. Simplify and solve for h(y). 4. Substitute the result for h(y) in the expression for φ from step 1 and then set φ = 0. This is the solution. Alternatively: 1. Let φ = � N(x, y)dy + g(x) 2. Set ∂φ ∂x = M(x, y) 3. Simplify and solve for g(x). 4. Substitute the result for g(x) in the expression for φ from step 1 and then set φ = 0. This is the solution. Integrating Factors Case 1: If P(x, y) depends only on x, where (18) P(x, y) = My − Nx N =⇒ µ(y) = e � P (x)dx then (19) µ(x)M(x, y)dx + µ(x)N(x, y)dy = 0 is exact. Case 2: If Q(x, y) depends only on y, where (20) Q(x, y) = Nx − My M =⇒ µ(y) = e � Q(y)dy Then (21) µ(y)M(x, y)dx + µ(y)N(x, y)dy = 0 is exact. 1� 2014 http://integral-table.com. This work is licensed under the Creative Commons Attribution – Noncommercial – No Derivative Works 3.0 United States License. To view a copy of this license, visit: http://creativecommons.org/licenses/by-nc-nd/3.0/us/. 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Second Order Linear Equations General Form of the Equation General Form: a(t)y′′ + b(t)y′ + c(t)y = g(t) (22) Homogeneous: a(t)y′′ + b(t)y′ + c(t)y = 0 (23) Standard Form: y′′ + p(t)y′ + q(t)y = f(t) (24) The general solution of (22) or (24) is (25) y = C1y1(t) + C2y2(t) + yp(t) where y1(t) and y2(t) are linearly independent solutions of (23). Linear Independence and The Wronskian Two functions f(x) and g(x) are linearly dependent if there exist numbers a and b, not both zero, such that af(x)+bg(x) = 0 for all x. If no such numbers exist then they are linearly inde- pendent. If y1 and y2 are two solutions of (23) then Wronskian: W(t) = y1(t)y′ 2(t) − y′ 1(t)y2(t) (26) Abel’s Formula: W(t) = Ce− � p(t)dt (27) and the following are all equivalent: 1. {y1, y2} are linearly independent. 2. {y1, y2} are a fundamental set of solutions. 3. W(y1, y2)(t0) ̸= 0 at some point t0. 4. W(y1, y2)(t) ̸= 0 for all t. Initial Value Problem (28)    y′′ + p(t)y′ + q(t)y = 0 y(t0) = y0 y′(t0) = y1 Linear Equation: Constant Coefficients Homogeneous: ay′′ + by′ + cy = 0 (29) Non-homogeneous: ay′′ + by′ + cy = g(t) (30) Characteristic Equation: ar2 + br + c = 0 (31) Quadratic Roots: r = −b ± √ b2 − 4ac 2a (32) The solution of (29) is given by: Real Roots(r1 ̸= r2) : yh = C1er1t + C2er2t (33) Repeated(r1 = r2) : yh = (C1 + C2t)er1t (34) Complex(r = α ± iβ) : yH = eαt(C1 cos βt + C2 sin βt) (35) The solution of (30) is y = yp + yh where yh is given by (33) through (35) and yp is found by undetermined coefficients or reduction of order. Heuristics for Undetermined Coefficients (Trial and Error) If f(t) = then guess that a particular solution yp = Pn(t) ts(A0 + A1t + · · · + Antn) Pn(t)eat ts(A0 + A1t + · · · + Antn)eat Pn(t)eat sin bt tseat[(A0 + A1t + · · · + Antn) cos bt or Pn(t)eat cos bt +(A0 + A1t + · · · + Antn) sin bt] Method of Reduction of Order When solving (23), given y1, then y2 can be found by solving (36) y1y′ 2 − y′ 1y2 = Ce− � p(t)dt The solution is given by (37) y2 = y1 � e− � p(x)dxdx y1(x)2 Method of Variation of Parameters If y1(t) and y2(t) are a fundamental set of solutions to (23) then a particular solution to (24) is (38) yP (t) = −y1(t) � y2(t)f(t) W(t) dt + y2(t) � y1(t)f(t) W(t) dt Cauchy-Euler Equation ODE: ax2y′′ + bxy′ + cy = 0 (39) Auxilliary Equation: ar(r − 1) + br + c = 0 (40) The solutions of (39) depend on the roots r1,2 of (40): Real Roots: y = C1xr1 + C2xr2 (41) Repeated Root: y = C1xr + C2xr ln x (42) Complex: y = xα[C1 cos(β ln x) + C2 sin(β ln x)] (43) In (43) r1,2 = α ± iβ, where α,β ∈ R Series Solutions (44) (x − x0)2y′′ + (x − x0)p(x)y′ + q(x)y = 0 If x0 is a regular point of (44) then (45) y1(t) = (x − x0)n ∞ � k=0 ak(x − xk)k At a Regular Singular Point x0: Indicial Equation: r2 + (p(0) − 1)r + q(0) = 0 (46) First Solution: y1 = (x − x0)r1 ∞ � k=0 ak(x − xk)k (47) Where r1 is the larger real root if both roots of (46) are real or either root if the solutions are complex.