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```FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS
14.28
Find the solution of
(5x + y − 7)
14.29
Find the solution y = y(x) of
x
14.30
dy
= 3(x + y + 1).
dx
dy
y2
+ y − 3/2 = 0,
dx
x
subject to y(1) = 1.
Find the solution of
(2 sin y − x)
14.31
if (a) y(0) = 0, and (b) y(0) = π/2.
Find the family of solutions of
d2 y
+
dx2
dy
dx
dy
= tan y,
dx
2
+
dy
=0
dx
that satisfy y(0) = 0.
14.1
14.3
14.5
14.7
14.9
14.11
14.13
14.15
14.17
14.19
14.21
14.23
14.25
14.27
14.29
N(t) = N0 exp(−λt).
(a) exact, x2 y 4 + x2 + y 2 = c; (b) IF = x−1/2 , x1/2 (x + y) = c; (c) IF =
sec2 x, y 2 tan x + y = c.
(a) IF = (1 − x2 )−2 , y = (1 − x2 )(k + sin−1 x); (b) IF = cosec x, leading to
y = k sin x + cos x; (c) exact equation is y −1 (dx/dy) − xy −2 = y, leading to
x = y(k + y 2 /2).
t
y(t) = e−t/α α−1 et /α f(t )dt ; (a) y(t) = 1 − e−t/α ; (b) y(t) = α−1 e−t/α ; (c) y(t) =
−t/α
−t/β
(e
−e
)/(α − β). It becomes case (b).
Note that, if the angle between the tangent and the radius vector is α, then
cos α = dr/ds and sin α = p/r.
Homogeneous equation, put y = vx to obtain (1 − v)(v 2 + 2v + 2)−1 dv = x−1 dx;
write 1 − v as 2 − (1 + 1v), and v 2 + 2v + 2 2as 1 + (1 + v)2 ;
A[x2 + (x + y)2 ] = exp 4 tan−1 [(x + y)/x] .
(1 + s)(dȳ/ds) + 2ȳ = 0. C = 1; use separation of variables to show directly that
y(t) = te−t .
The equation is of the form of (14.22), set v = x + y; x + 3y + 2 ln(x + y − 2) = A.
The equation is isobaric with weight y = −2; setting y = vx−2 gives
v −1 (1 − v)−1 (1 − 2v) dv = x−1 dx; 4xy(1 − x2 y) = 1.
which has solution x = (p−1)−2 ,
The curve must satisfy√y = (1−p−1 )−1 (1−x+px),
√
leading to y = (1 ± x)2 or x = (1 ± y)2 ; the singular solution p = 0 gives
straight lines joining (θ, 0) and (0, 1 − θ) for any θ.
v = qu + q/(q − 1), where q = dv/du. General solution y 2 = cx2 + c/(c − 1),
hyperbolae for c > 0 and ellipses for c < 0. Singular solution y = ±(x ± 1).
(a) Integrating factor is (a2 + x2 )1/2 , y = (a2 + x2 )/3 + A(a2 + x2 )−1/2 ; (b) separable,
y = x(x2 + Ax + 4)−1 .
Use Laplace transforms; x̄s(s2 + 4) = s + s2 − 2e−3s ;
x(t) = 12 sin 2t + cos 2t − 12 H(t − 3) + 12 cos(2t − 6)H(t − 3).
This is Clairaut’s equation
√ with F(p) = A/p. General solution y = cx + A/c;
singular solution, y = 2 Ax.
Either Bernoulli’s equation with n = 2 or an isobaric equation with m = 3/2;
y(x) = 5x3/2 /(2 + 3x5/2 ).
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