Separable Differential Equations
𝑑𝑦
Def. A first order differential equation, = 𝐻(𝑥, 𝑦), is called separable if we
𝑑𝑥
can write 𝐻 (𝑥, 𝑦) = 𝑔(𝑥 )ℎ(𝑦).
In that case:
𝑑𝑦
= 𝑔(𝑥)ℎ(𝑦)
𝑑𝑥
1
𝑑𝑦 = 𝑔(𝑥 )𝑑𝑥
ℎ(𝑦)
then,
1
∫ ℎ(𝑦) 𝑑𝑦 = ∫ 𝑔(𝑥 )𝑑𝑥 .
Ex. Solve the initial value problem
𝑑𝑦
= −8𝑥𝑦; 𝑦(0) = 4.
𝑑𝑥
Notice that 𝐻 (𝑥, 𝑦) = (−8𝑥 )(𝑦) = 𝑔(𝑥 )ℎ(𝑦).
𝑑𝑦
= −8𝑥𝑑𝑥
𝑦
𝑑𝑦
∫ 𝑦 = ∫ −8𝑥𝑑𝑥
ln| 𝑦| + 𝑐1 = −4𝑥 2 + 𝑐2
ln| 𝑦| = −4𝑥 2 + 𝑐3
2
𝑒 ln| 𝑦| = 𝑒 −4𝑥 +𝑐3
2 2
|𝑦| = 𝑒 −4𝑥 ⋅ 𝑒 𝑐3 = 𝐴𝑒 −4𝑥 ; (general solution).
, 2
𝑦(0) = 4 so,
2
|𝑦(0)| = 4 = 𝐴𝑒 −4(0)
4 = 𝐴𝑒 0 = 𝐴
2
So, |𝑦| = 4𝑒 −4𝑥
But 𝑦(0) = 4 > 0 so 𝑦 > 0 near 𝑥 = 0.
Thus 𝑦 = |𝑦| and
2
𝑦 = 4𝑒 −4𝑥 (particular solution).
Ex. Solve the initial value problem
𝑑𝑦 5−4𝑥
= 𝑦(4𝑦 2 +2) ; 𝑦(1) = 2.
𝑑𝑥
𝑑𝑦 5−4𝑥 1
= 4𝑦 3 +2𝑦 = (5 − 4𝑥) (4𝑦3 +2𝑦) = 𝑔(𝑥)ℎ(𝑦).
𝑑𝑥
(4𝑦 3 + 2𝑦)𝑑𝑦 = (5 − 4𝑥 )𝑑𝑥
∫(4𝑦 3 + 2𝑦) 𝑑𝑦 = ∫(5 − 4𝑥 )𝑑𝑥
𝑦 4 + 𝑦 2 + 𝑐1 = 5𝑥 − 2𝑥 2 + 𝑐2
𝑦 4 + 𝑦 2 = 5𝑥 − 2𝑥 2 + 𝑐3 (general solution).
One can’t easily solve this equation for 𝑦 in terms of 𝑥, so we leave the
solution in this form. This equation represents a set of curves in the 𝑥-𝑦 plane
𝑑𝑦 5−4𝑥
where = 4𝑦 3 +2𝑦 at every point (𝑥, 𝑦) that fits the equation
𝑑𝑥
𝑦 4 + 𝑦 2 = 5𝑥 − 2𝑥 2 + 𝑐3 .