DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS NON-LINEAR LINEAR (in y) LINEAR W/ CST COEFFs (in y) FIRST- ORDER 4(y ) 2 +x cos y = x 2 4x 2 y + y cos x = x 2 4y + 3y = cos x ORDINARY DIFF EQs SECOND- ORDER 4y y + xy 1 2 = x 2 4x 2 y + yx 1 2 = x 2 4y + 2y + 3y = x 2 HIGHER- ORDER y (5) y... cos y... y 1/2 y (5)... y... y PARTIAL DIFF EQs y x 1 y x n 4 y x 1 + 3 y x 2 = x 1 x 2

Book: Kreyszig, 10 th ed. general explicit: y = f x, y P x,y = Q x,y P x, y + Q x, y = 0 p4 (1.1) linear: if yy + p x y = q x p27 (1.5)

general explicit: y = f x, y P x,y = Q x,y P x, y + Q x, y = 0 p4 (1.1) p12 (1.3) separable: if P x + Q y = 0 then integrate separately and resolve for y

general explicit: y = f x, y P x,y = Q x,y P x, y + Q x, y = 0 p4 (1.1) p12 (1.3) separable: if P x + Q y = 0 then integrate separately and resolve for y linear substitution: if = f(aa + bb + c) then use u = aa + bb + c and solve separable eq. = a+bb(u)

general explicit: y = f x, y P x,y = Q x,y P x, y + Q x, y = 0 p4 (1.1) separable: if P x + Q y = 0 then integrate separately and resolve for y linear substitution: if = f(aa + bb + c) then use u = aa + bb + c and solve separable eq. = a+bb(u) p12 (1.3) simple balanced case: if y = f y x, that is P(x, y) Q(x, y) = f(y x) then use y = xv p17 (mid)

general explicit: y = f x, y P x,y = Q x,y P x, y + Q x, y = 0 p4 (1.1) separable: if P x + Q y = 0 then integrate separately and resolve for y linear substitution: if = f(aa + bb + c) then use u = aa + bb + c and solve separable eq. = a+bb(u) p12 (1.3) simple balanced case: if y = f y x, that is P(x, y) Q(x, y) = f(y x) then use y = xv generally balanced ( homogeneous degree): if P tx, ty = t α P and Q tt, tt = t β Q and α = β (same degree of homogeneity) then use y = xx and solve 1 x α P + vq + xxxx = 0 i.e. P(1, v) + vv(1, v) + xx(1, v)dv = 0 i.e. the separable eq. x = P 1,v + v 1 Q 1,v p17 (mid)

general explicit: y = f x, y P x,y = Q x,y P x, y + Q x, y = 0 p4 (1.1) separable: if P x + Q y = 0 then integrate separately and resolve for y linear substitution: if = f(aa + bb + c) then use u = aa + bb + c and solve separable eq. = a+bb(u) p12 (1.3) simple balanced case: if y = f y x, that is P(x, y) Q(x, y) = f(y x) then use y = xv generally balanced ( homogeneous degree): if P tx, ty = t α P and Q tt, tt = t β Q and α = β (same degree of homogeneity) then use y = xx and solve 1 x α P + vq + xxxx = 0 i.e. P(1, v) + vv(1, v) + xx(1, v)dv = 0 i.e. the separable eq. x = P 1,v + v 1 Q 1,v p17 (mid) if (x,y) y exact: = Q(x,y) x p20 (1.4) then find F x, y such that F F = P and = Q, y integrate separately, and solution is F(x, y) = C

if (x,y) y exact: = Q(x,y) x p20 (1.4) then find F x, y such that F F = P and = Q, y integrate separately, and solution is F(x, y) = C exact with integrating factor : if Q y x then find μ(x) such that μp = μq : possible if 1 Q then F x, y such that F F = μp and = μμ F(x, y) = C y p23 (bott) = g(x) only,

if (x,y) y exact: = Q(x,y) x p20 (1.4) then find F x, y such that F F = P and = Q, y integrate separately, and solution is F(x, y) = C exact with integrating factor : if Q y x then find μ(x) such that μp = μq : possible if 1 Q then F x, y such that F F = μp and = μμ F(x, y) = C y or conversely, μ(y) exists if 1 P P Q y x p23 (bott) = g(x) only, = h(y) only

if (x,y) y exact: = Q(x,y) x p20 (1.4) then find F x, y such that F F = P and = Q, y integrate separately, and solution is F(x, y) = C exact with integrating factor : if Q y x then find μ(x) such that μp = μq : possible if 1 Q then F x, y such that F F = μp and = μμ F(x, y) = C y or conversely, μ(y) exists if 1 P P Q y x p23 (bott) = g(x) only, = h(y) only or try μ x, y = x m y n and solve μp = μq for m and n (no general condition)

FIRST-ORDER LINEAR ODEs linear: if yy + p x y = q x p27 (1.5) homogeneous linear: p28 (top) if + p x y = 0 ( complementary equation ) then: ln y c = p x + C y c = C, = natural regime α where α = eee p x is the integrating factor

FIRST-ORDER LINEAR ODEs linear: if yy + p x y = q x p27 (1.5) homogeneous linear: p28 (top) if + p x y = 0 ( complementary equation ) then: ln y c = p x + C y c = C, = natural regime α where α = eee p x is the integrating factor the integrating factor α x can also be found by setting: α d αy + αp x y = = αq x dα = α x p x with α, integrate where y p = 1 α d αα = αα y = y p + y c, α(x)q x is a particular solution p28 (bott) = forced regime

FIRST-ORDER LINEAR ODEs linear: if yy + p x y = q x p27 (1.5) homogeneous linear: p28 (top) if + p x y = 0 ( complementary equation ) then: ln y c = p x + C y c = C, = natural regime α where α = eee p x is the integrating factor the integrating factor α x can also be found by setting: α d αy + αp x y = = αq x dα = α x p x with α, integrate where y p = 1 α d αα = αα y = y p + y c, α(x)q x is a particular solution equivalent method: variation of the constant (find a particular solution by varying the coeff. of the homog. solution) use y p = C(x) α and solve = αα p28 (bott) = forced regime

FIRST-ORDER LINEAR ODEs linear: if yy + p x y = q x p27 (1.5) homogeneous linear: p28 (top) if + p x y = 0 ( complementary equation ) then: ln y c = p x + C y c = C, = natural regime α where α = eee p x is the integrating factor Bernoulli p31 (bott) if + p x y = q x yn then use u x = y 1 N and solve linear eq. + (1 N)p x u = (1 N)q(x) the integrating factor α x can also be found by setting: α d αy + αp x y = = αq x dα = α x p x with α, integrate where y p = 1 α d αα = αα y = y p + y c, α(x)q x is a particular solution equivalent method: variation of the constant (find a particular solution by varying the coeff. of the homog. solution) use y p = C(x) α and solve = αα p28 (bott) = forced regime