This cheat sheet summarizes techniques for solving first and second order ordinary differential equations. It provides the main methods for solving trivial, linear non-homogeneous, and separable first order differential equations, as well as finding the complementary and particular solutions for second order differential equations with constant coefficients. Students are advised to memorize and practice these problem-solving approaches.

1. DIFFERENTIAL EQUATIONS MA133 CHEAT SHEET
This is a summary sheet for the first half of the module, solving first and second
order ordinary differential equations. We assume conditions for FTC hold. You
should learn all these techniques by heart, and practice, practice, practice!
First Order Differential Equations We consider the main scenarios
Trivial Case (Section 1.1)
dx
= f (t)
dt
By Fundamental Theorem of Calculus simply integrate both sides with
respect to t
x(t) = f (t) dt
Linear Non-homogeneous (Sections 1.3/1.4/1.5)
dx
+ p(t)x = q(t)
dt
Multiply both sides by an Integrating Factor P (t) = exp ( p(t)dt) so that
d
(P (t)x(t)) = P (t)q(t)
dt
Then integrate so that
x(t) = P (t)−1 P (s)q(s)ds + AP (t)−1
t
Separable Equations (Section 1.6)
dx
= f (x)g(t)
dt
First look for constant solutions, i.e. where f (x) = 0. Then look for non-
constant solutions (so f (x) never zero) and ”divide both sides by f (x),
multiply both sides by dt and integrate”.
dx
= g(t)dt
f (x)
Autonomous First Order ODEs (Section 1.9)
dx
= f (x)
dt
Look for fixed points x∗ , which satisfy f (x∗ ) = 0, i.e. are points where
dx
= 0. A fixed point x∗ is stable if f ′ (x∗ ) < 0 and unstable if f ′ (x∗ ) > 0.
dt
1

2. Second Order Ordinary Differential Equations With Constant Coefficients
d2 x dx
+ b + cx = f (t)
a 2
dt dt
The solution consists of x(t) = xc (t)+xp (t) where xc (t), the complementary so-
lution, solves the homogeneous case f (t) = 0 and xp (t), the particular integral,
gives the f (t).
The Complementary Solution Solves
d2 x dx
+ b + cx = 0
a 2
dt dt
Find the roots to the auxiliary equations
aλ2 + bλ + c = 0
√
−b± b2 −4ac
i.e. λ± = then we have
2a
• Real roots k1 , k2 complementary solution is
Aek1 t + Bek2 t
• Repeated real root k complementary solution is
Aekt + Btekt
• Complex roots p ± iq complementary solution is
ept (A sin(qt) + B cos(qt))
or
Aept cos(qt − φ)
The Particular Integral Functions to ”guess”:
f (t) Try solution xp (t) =
aekt (k not a root) Aekt
aekt (k a root) Atekt or At2 ekt
a sin(ωt) or a cos(ωt) A sin(ωt) + B cos(ωt)
atn where n ∈ N P (t) general polynomial of degree n
atn ekt P (t)ekt , P (t) general polynomial of degree n
n
t (a sin(ωt) + b cos(ωt) P1 (t) sin(ωt) + P2 (t) cos(ωt)
where Pi (t) general polynomial of degree n
ekt (a sin(ωt) + b cos(ωt)) ekt (A sin(ωt) + B cos(ωt))
2