Math 308 Section-Zhou,
Review for Exam 1
1. 1st-order linear
Factor Method. y ′ + p(t)y = g(t), y(t0 ) = y0 .
∫ D.E.: Integrating
∫
p(t)dt
1
Set µ(t) = e
, y(t) = µ(t)
[ µ(t)g(t)dt + C]. Then use I.C. y(t0 ) = y0 to determine C.
2. 1st-order Separable D.E.:
dy
dx



 g(x)/f (t)
=
∫
g(x)f (y) ⇒


 f (y)/g(x)
∫
f (y)dy =
g(x)dx
∫ dy
∫
g(x)dx
f (y) =
∫
∫
dy
f (y)
dx
=
g(x)
Then solve the two integrals. The resulting equation defines a solution y = y(x) implicitly.
3. 1st-order Exact D.E.: M (x, y)dx + N (x, y)dy = 0. Check My = Nx . If yes, it is exact.
Use Fx = M, Fy = N to find F (x, y), or solve the integrals and determine α(y), β(x) from
∫
F (x, y) =
∫
M (x, y)dx + α(y) =
N (x, y)dy + β(x),
or their combinations. y = y(x) is defined by F (x, y) = C where C is determined if I.C. is given.
4. 1st-order D.E. Application: (a) Mixing problem.
(b) Loan problem. dS(t)
dt = rS − 12k, S(0) = S0 .
dQ(t)
dt
= cr −
Q
W r, Q(0)
= Q0 ,
5. Autonomous D.E. y ′ = f (y), y(0) = y0 . Critical point: f (y) = 0, (in)stability (AS, US).
y
y
(a) f (y) = ±r(1 − K
)y; (b) f (y) = ±r(1 − Ty )(1 − K
)y, 0 < T < K.
Draw f (y) vs y (+, −, direction of y) and y(t) vs t (asymptote, threshold) for different y0 .
6. 2nd-order D.E. with constant coefficients. Hom: ay ′′ + by ′ + cy = 0. Solve λ1,2 = −b±
(a) b2 − 4ac > 0, λ1 ̸= λ2 , real, distinct y(t) = c1 eλ1 t + c2 eλ2 t ;
b
(b) b2 = 4ac, λ1 = λ2 , real, repeated, y(t) =√ e− 2a t (c1 + c2 t);
2
b
(c) b2 − 4ac < 0, λ = µ ± iν, µ = − 2a
, ν = 4ac−b
, y(t) = eµt (c1 sin(νt) + c2 cos(νt)).
2a
√
b2 −4ac
.
2a
′
7. Reduction of Order. y ′′ + p(t)y
∫ + q(t)y = 0. Given y1 ̸= 0, find y2 (t) = v(t)y1 (t) where
v(t) =
∫
u(t)dt and u(t) =
e
−
p(t)dt
[y1 (t)]2
. y(t) = y1 (t)(c1 + c2 v(t)).
8. y ′′ + p(t)y ′ + q(t)y = g(t). Superposition Principle: y(t) = c1 y1 (t) + c2 y2 (t) + yp (t) where
c1 y1 (t) + c2 y2 (t) is the general solution to the corresponding hom D.E. y ′′ + p(t)y ′ + q(t)y = 0 and yp is a
particular solution to y ′′ + p(t)y ′ + q(t)y = g(t).
9. To find yp to ay ′′ + by ′ + cy = g(t). The Method of Undetermined Coefficients. s = 0, 1, 2?
When g(t) = pn (t) = an tn + · · · + a1 t + a0 , pn (t)eαt , pn (t)eαt (a sin(βt) + b cos(βt), then
yp (t) = ts (An tn + · · · + A0 ), ts (An tn + · · · + A0 )eαt , ts eαt [(An tn + · · · + A0 ) sin(βt) + (Bn tn + · · · + B0 ) cos(βt)]
where s = 0, 1, 2 is chosen so that no term in yp is a solution to ay ′′ + by ′ + cy = 0.
10. The Method of Variation of Parameters. Given fundamental set of solutions y1 , y2 to y ′′ +p(t)y ′ +q(t)y = 0.
Then a particular solution to y ′′ + p(t)y ′ + q(t)y = g(t) is given by yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t) where
∫
v1 (t) =
−y2 (t)g(t)
dt, v2 (t) =
W [y1 , y2 ](t)
∫
y (t) y (t)
y1 (t)g(t)
2
1
dt, W [y1 , y2 ](t) = ′
y1 (t) y2′ (t)
W [y1 , y2 ](t)
y(t) = c1 y1 (t) + c2 y2 (t) + yp (t).
1
= y1 (t)y2′ (t) − y2 (t)y1′ (t).