Mathematical Methods for Physics and Engineering - Matematica.NET

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS14.28 Find the solution of(5x + y − 7) dy =3(x + y +1).dx14.29 Find the solution y = y(x) ofx dydx + y − y2=0,x3/2 subject to y(1) = 1.14.30 Find the solution of(2 sin y − x) dydx =tany,if (a) y(0) = 0, and (b) y(0) = π/2.14.31 Find the family of solutions ofthat satisfy y(0) = 0.( )d 2 2y dydx + + dy2 dx dx =014.5 Hints and answers14.1 N(t) =N 0 exp(−λt).14.3 (a) exact, x 2 y 4 + x 2 + y 2 = c; (b) IF = x −1/2 , x 1/2 (x + y) = c; (c) IF =sec 2 x, y 2 tan x + y = c.14.5 (a) IF = (1 − x 2 ) −2 , y =(1− x 2 )(k +sin −1 x); (b) IF = cosec x, leading toy = k sin x +cosx; (c) exact equation is y −1 (dx/dy) − xy −2 = y, leading tox = y(k + y 2 /2).14.7 y(t) =e ∫ −t/α t α −1 e t′ /α f(t ′ )dt ′ ;(a)y(t) =1− e −t/α ;(b)y(t) =α −1 e −t/α ;(c)y(t) =(e −t/α − e −t/β )/(α − β). It becomes case (b).14.9 Note that, if the angle between the tangent and the radius vector is α, thencos α = dr/ds and sin α = p/r.14.11 Homogeneous equation, put y = vx to obtain (1 − v)(v 2 +2v +2) −1 dv = x −1 dx;write 1 − v as 2 − (1 + v), and v 2 +2v +2as1+(1+v) 2 ;A[x 2 +(x + y) 2 ]=exp { 4tan −1 [(x + y)/x] } .14.13 (1 + s)(dȳ/ds)+2ȳ =0.C = 1; use separation of variables to show directly thaty(t) =te −t .14.15 The equation is of the form of (14.22), set v = x + y; x +3y +2ln(x + y − 2) = A.14.17 The equation is isobaric with weight y = −2; setting y = vx −2 givesv −1 (1 − v) −1 (1 − 2v) dv = x −1 dx; 4xy(1 − x 2 y)=1.14.19 The curve must satisfy y =(1−p −1 ) −1 (1−x+px), which has solution x =(p−1) −2 ,leading to y =(1± √ x) 2 or x =(1± √ y) 2 ; the singular solution p ′ = 0 givesstraight lines joining (θ, 0) and (0, 1 − θ) foranyθ.14.21 v = qu + q/(q − 1), where q = dv/du. General solution y 2 = cx 2 + c/(c − 1),hyperbolae for c>0 and ellipses for c