Differential Equations Study Guide1 First Order Equations dy = f (x, y) dx Initial Value Problem: y 0 = f (x, y), y(x0 ) = y0 General Form of ODE: (1) (2) Linear Equations General Form: y 0 + p(x)y = f (x) (3) (4) (5) (6) Method for Solving Exact Equations: R 1. Let φ = M (x, y)dx + h(y) R p(x)dx Integrating Factor: µ(x) = e d (µ(x)y) = µ(x)f (x) =⇒ dx Z 1 General Solution: y = µ(x)f (x)dx + C µ(x) Homogeneous Equations ∂φ = N (x, y) ∂y 3. Simplify and solve for h(y). 2. Set 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: R 1. Let φ = N (x, y)dy + g(x) ∂φ = M (x, y) ∂x 3. Simplify and solve for g(x). 2. Set (7) (8) General Form: y 0 = f (y/x) Substitution: y = zx 4. Substitute the result for g(x) in the expression for φ from step 1 and then set φ = 0. This is the solution. =⇒ y 0 = z + xz 0 (9) The result is always separable in z: dx dz = f (z) − z x (10) Integrating Factors Case 1: If P (x, y) depends only on x, where Bernoulli Equations (11) (12) 0 General Form: y + p(x)y = q(x)y n Substitution: z = y 1−n (18) P (x, y) = R M y − Nx =⇒ µ(y) = e P (x)dx N then The result is always linear in z: (13) z 0 + (1 − n)p(x)z = (1 − n)q(x) (19) µ(x)M (x, y)dx + µ(x)N (x, y)dy = 0 is exact. Exact Equations Case 2: If Q(x, y) depends only on y, where (14) (20) (15) (16) (17) General Form: M (x, y)dx + N (x, y)dy = 0 ∂M ∂N Text for Exactness: = ∂y ∂x Solution: φ = C where ∂φ ∂φ M= and N = ∂x ∂y Q(x, y) = R Nx − M y =⇒ µ(y) = e Q(y)dy M 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 Heuristics for Undetermined Coefficients (Trial and Error) (22) General Form: a(t)y 00 + b(t)y 0 + c(t)y = g(t) (23) Homogeneous: a(t)y 00 + b(t)y 0 + c(t)y = 0 (24) Standard Form: y 00 + p(t)y 0 + q(t)y = f (t) If f (t) = Pn (t) Pn (t)eat Pn (t)eat sin bt or Pn (t)eat cos bt then guess that a particular solution yp = ts (A0 + A1 t + · · · + An tn ) ts (A0 + A1 t + · · · + An tn )eat ts eat [(A0 + A1 t + · · · + An tn ) cos bt +(A0 + A1 t + · · · + An tn ) sin bt] The general solution of (22) or (24) is (25) y = C1 y1 (t) + C2 y2 (t) + yp (t) Method of Reduction of Order where y1 (t) and y2 (t) are linearly independent solutions of (23). Linear Independence and The Wronskian When solving (23), given y1 , then y2 can be found by solving (36) y1 y20 − y10 y2 = Ce− R p(t)dt The solution is given by Two functions f (x) and g(x) are linearly dependent if there Z − R p(x)dx e dx exist numbers a and b, not both zero, such that af (x) + bg(x) = 0 (37) y2 = y1 2 for all x. If no such numbers exist then they are linearly indey1 (x) pendent. If y1 and y2 are two solutions of (23) then (26) (27) Wronskian: W (t) = Method of Variation of Parameters y1 (t)y20 (t) − R − p(t)dt y10 (t)y2 (t) Abel’s Formula: W (t) = Ce and the following are all equivalent: 1. {y1 , y2 } are linearly independent. 2. {y1 , y2 } are a fundamental set of solutions. If y1 (t) and y2 (t) are a fundamental set of solutions to (23) then a particular solution to (24) is Z Z y2 (t)f (t) y1 (t)f (t) (38) yP (t) = −y1 (t) dt + y2 (t) dt W (t) W (t) Cauchy-Euler Equation 3. W (y1 , y2 )(t0 ) 6= 0 at some point t0 . 4. W (y1 , y2 )(t) 6= 0 for all t. ODE: ax2 y 00 + bxy 0 + cy = 0 (40) Auxilliary Equation: ar(r − 1) + br + c = 0 The solutions of (39) depend on the roots r1,2 of (40): Initial Value Problem (28) (39) (41) 00 y + p(t)y 0 + q(t)y = 0 y(t0 ) = y0 0 y (t0 ) = y1 Real Roots: y = C1 xr1 + C2 xr2 (42) Repeated Root: y = C1 xr + C2 xr ln x (43) Complex: y = xα [C1 cos(β ln x) + C2 sin(β ln x)] In (43) r1,2 = α ± iβ, where α,β ∈ R Linear Equation: Constant Coefficients Series Solutions (29) (30) (31) (32) 00 0 Homogeneous: ay + by + cy = 0 Non-homogeneous: ay 00 + by 0 + cy = g(t) Characteristic Equation: ar2 + br + c = 0 √ −b ± b2 − 4ac Quadratic Roots: r = 2a (44) If x0 is a regular point of (44) then (45) Real Roots(r1 6= r2 ) : yh = C1 er1 t + C2 er2 t (34) Repeated(r1 = r2 ) : yh = (C1 + C2 t)er1 t (35) Complex(r = α ± iβ) : yH = eαt (C1 cos βt + C2 sin βt) y1 (t) = (x − x0 )n ∞ X ak (x − xk )k k=0 The solution of (29) is given by: (33) (x − x0 )2 y 00 + (x − x0 )p(x)y 0 + q(x)y = 0 At a Regular Singular Point x0 : (46) (47) Indicial Equation: r2 + (p(0) − 1)r + q(0) = 0 ∞ X First Solution: y1 = (x − x0 )r1 ak (x − xk )k k=0 The solution of (30) is y = yp + yh where yh is given by (33) through (35) and yp is found by undetermined coefficients or Where r1 is the larger real root if both roots of (46) are real or reduction of order. either root if the solutions are complex.

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