Separable
Form
𝑔(𝑦)𝜕𝑦 = 𝑓(𝑥)𝜕𝑥
Solution
∫ 𝑔(𝑦)𝜕𝑦 = ∫ 𝑓(𝑥)𝜕𝑥
Linear
Form
𝑎1 (𝑥)
𝜕𝑦
+ 𝑎0 (𝑥)𝑦 = 𝑔(𝑥)
𝜕𝑥
𝑓(𝑥)𝑦′ + 𝑔(𝑥)𝑦 = ℎ(𝑥)
𝜕𝑦
+ 𝑃 (𝑥)𝑦 = 𝑄(𝑥)
𝜕𝑥
Solution
𝑦=
∫ 𝜇(𝑥)𝑄(𝑥)𝜕𝑥
𝜇(𝑥)
where
𝜇(𝑥) = 𝑒∫ 𝑃 (𝑥)𝜕𝑥
Exact
Form
𝑀 (𝑥, 𝑦)𝜕𝑥 + 𝑁 (𝑥, 𝑦)𝜕𝑦 = 0
Check
𝜕𝑀
𝜕𝑁
=
𝜕𝑦
𝜕𝑥
Solution
𝜕𝐹 (𝑥, 𝑦) = 𝑀 (𝑥, 𝑦)𝜕𝑥 + 𝑁 (𝑥, 𝑦)𝜕𝑦
𝐹 (𝑥, 𝑦) = ∫ 𝑀 (𝑥, 𝑦)𝜕𝑥 + 𝑔(𝑦)
where
𝑔′ (𝑦) =
𝜕𝐹
𝜕
−
∫ 𝑀 (𝑥, 𝑦)𝜕𝑥
𝜕𝑦
𝜕𝑦
= 𝑁 (𝑥, 𝑦) −
𝜕
∫ 𝑀 (𝑥, 𝑦)𝜕𝑥
𝜕𝑦
𝑔(𝑦) = ∫(𝑁 (𝑥, 𝑦) −
𝜕
∫ 𝑀 (𝑥, 𝑦)𝜕𝑥)𝜕𝑦
𝜕𝑦
Special Integrating Factors
Given
𝜕𝑀
− 𝜕𝑁
𝜕𝑦
𝜕𝑥
⎡
𝜇(𝑥) = exp⎢∫
𝜕𝑥⎤
⎥
𝑁
⎣
⎦
or
𝜕𝑁
− 𝜕𝑀
𝜕𝑥
𝜕𝑦
𝜇(𝑦) = exp⎡
∫
𝜕𝑦⎤
⎢
⎥
𝑀
⎣
⎦
then
𝜇(𝑥)𝑀 (𝑥, 𝑦)𝜕𝑥 + 𝜇(𝑥)𝑁 (𝑥, 𝑦)𝜕𝑦 = 0
(or the equivalent with 𝜇(𝑦)) is exact.
Homogeneous Equations
Form
𝜕𝑦
𝑦
= 𝐺( )
𝜕𝑥
𝑥
Solution
Given 𝑣 = 𝑥𝑦
∫
1
1
𝜕𝑣 = ∫ 𝜕𝑥
𝑥
𝐺(𝑣) − 𝑣
then substitute 𝑥𝑦 for 𝑣.
Form 𝐺(𝑎𝑥 + 𝑏𝑦)
Form
𝜕𝑦
= 𝐺(𝑎𝑥 + 𝑏𝑦)
𝜕𝑥
Solution
Given 𝑧 = 𝑎𝑥 + 𝑏𝑦
∫
then substitute 𝑎𝑥 + 𝑏𝑦 for 𝑧.
Bernoulli
Form
1
𝜕𝑧 = ∫ 𝜕𝑥
𝑎 + 𝑏𝐺(𝑧)
𝜕𝑦
+ 𝑃 (𝑥)𝑦 = 𝑄(𝑥)𝑦𝑛
𝜕𝑥
Solution
Given 𝑣 = 𝑦1−𝑛
𝜕𝑣
+ (1 − 𝑛)𝑃 (𝑥)𝑣 = (1 − 𝑛)𝑄(𝑥)
𝜕𝑥
After solving using the linear method, substitute 𝑦1−𝑛 for 𝑣.
Equations with Linear Coefficients
Form
(𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 )𝜕𝑥 + (𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 )𝜕𝑦 = 0
Solution
Given 𝑥 = 𝑢 + ℎ and 𝑦 = 𝑣 + 𝑘 such that
𝑎1 ℎ + 𝑏1 𝑘 + 𝑐1 = 0,
𝑎2 ℎ + 𝑏2 𝑘 + 𝑐2 = 0
then
𝜕𝑣
𝑎 + 𝑏1 (𝑣/𝑢)
=− 1
𝜕𝑢
𝑎2 + 𝑏2 (𝑣/𝑢)
Solve as a homogeneous equation