Bernoulli Equation
The general first order differential equation of the form
𝑑𝑦
𝑑𝑥
+ P(x) y = Q(x)𝑦𝑛 ,
n≠ 0, 1
To solve : divide LHS and RHS by 𝑦𝑛
𝑑𝑦
1−𝑛
+
P(x)
𝑦
= Q(x)
𝑑𝑥
= 𝑦1−𝑛 , dv/dx = (1-n ) 𝑦 −𝑛
𝑦 −𝑛
we let v
dy/dx
The equation reduces to form
𝑑𝑣
+
𝑑𝑥
(1-n) P(x) v = (1-n) Q(x),
next find Integrating factor u(x) = 𝑒 (1−𝑛) ‫ 𝑥𝑑)𝑥(𝑃 ׬‬and times on LHS and RHS
Simplify we get : v𝑒 (1−𝑛) ‫( = 𝑥𝑑)𝑥(𝑃 ׬‬1-n) ‫(׬ 𝑒𝑄 ׬‬1−𝑛)𝑃(𝑥)𝑑𝑥 𝑑𝑥 + C.
Tutorial Q5: Solve x
𝑑𝑦
𝑑𝑥
+ y = x 𝑦3
add this as (b) on tutorial
ClAIRAUT’s EQUATION
Solve y = p x ± 𝑝2 + 1,
𝑤ℎ𝑒𝑟𝑒 𝑝 = 𝑑𝑦/𝑑𝑥
Differentiating y
𝑑𝑦
𝑑𝑥
𝑑𝑝
𝑑𝑥
=p=p+x
x ±
𝑑𝑝
𝑝
𝑑𝑝
± 2
𝑑𝑥
𝑝 +1 𝑑𝑥
𝑝
𝑝2 +1
=0
𝑑𝑝
Case 1 : 𝑑𝑥 = 0 , so p = c , hence the general solution is y = c x ± 𝑐 2 + 1,
Case2 : x ±
y=∓
𝑝
𝑝2 +1
= 0 , so x = ∓
𝑝2
±
𝑝2 +1
𝑝
, hence
𝑝2 +1
𝑝2 + 1 = ∓
1
𝑝2 +1
To eliminate p
𝑥 2 + 𝑦2 = 1, satisfies the above Eq. but cannot be solve for not all c, singular solution
Equation of order higher than one
𝑑 2𝑦
𝑑𝑥2
Solve
y is missing in the equation. Let p =
𝑑𝑦
+ 2 𝑑𝑥 = 4 x
𝑑𝑦
𝑑𝑥
= y’ , so p’ = y”
The above equation becomes p’ + 2 p = 4 x
thus we find Integrating factor u = exp ( ‫ ׬‬2 𝑑𝑥) = 𝑒 2𝑥
multiply both LHS and RHS:
𝑑(𝑝𝑒 2𝑥 )
𝑑𝑥
p’ 𝑒 2𝑥 + 2p 𝑒 2𝑥 = 4 x 𝑒 2𝑥
= 4 x 𝑒 2𝑥
p 𝑒 2𝑥 = 4 ‫ ׬‬x 𝑒 2𝑥 dx + C
p = 2x – 1 + C 𝑒 −2𝑥
To get back y from y = p’ , we integrate again
y = 𝑥 2 - x + A 𝑒 −2𝑥 + D
RECALL
Example:
Find the solution of the initial value problem (IVP):
𝑑𝑦
𝑑𝑡
1
+ y = 1+ 𝑡 2
Solution: Using similar method
y(2) = 3;
(i) the integrating function is u(t) = 𝑒 𝑡,
(ii) since it has initial value. we employ the definite integral with appropriate limits
𝑑
[
𝑑𝑡
𝑒𝑡
𝑦𝑒 𝑡 ] = 1+ 𝑡 2
𝑡 𝑑
‫׬‬2 𝑑𝑠 [𝑦𝑒 𝑠 ]𝑑𝑠
𝑒 𝑡 𝑦(t)
=
– y(2)
𝑡 𝑒𝑠
‫׬‬2 1+𝑠2
𝑒2
y= 𝑒
−𝑡
=
ds
𝑒 𝑡𝑦(t)
2
[3𝑒 +
–3
𝑒2
𝑡 𝑒𝑠
‫׬‬2 1+𝑠2
=
ds]
𝑡 𝑒𝑠
‫׬‬2 1+𝑠2
ds
Exact first-order differential equation
Generally a differential equation is in the form
𝑑
Φ(t,y(t)) =
𝑑𝑡
0,
Φ(t, y(t)) = constant
solved by simply integrating giving
variable ‘y’ which is a function of ‘t’.
𝑑
Remark: Please note that, it is not always we can re-arrange a differential equation in the form 𝑑𝑡 Φ(t, y) = 0.
But looking at the “chain rule” of a partial differentiation
𝑑
Φ(t,
𝑑𝑡
Set
y) =
𝑑
Φ(t, y)
𝑑𝑡
𝜕Φ
𝜕𝑡
+
𝜕Φ 𝑑𝑦
𝜕𝑦 𝑑𝑡
= M(t,y) + N(t,y)
we identify M(t,y) =
𝜕Φ
𝜕𝑡
𝑑𝑦
𝑑𝑡
= 0, which we required !!!!
and N(t,y) =
𝜕Φ
.
𝜕𝑦
Definition:
The differential M(t,y) dt + N(t,y) dy = 0, is said to be exact if and only if
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
.
𝜕𝑡
Method to solve exact differential equation
Method (i)
Step 1: Knowing the equation M(t, y) =
𝜕Φ
,
𝜕𝑡
we determine Φ(t,y) up to arbitrary function of y along that
Φ(t, y) = ‫𝑡 𝑀 ׬‬, 𝑦 𝑑𝑡 + h(y).
Step2: The h(y) is then determined by h’(y) = N(t,y) - ‫׬‬
𝜕𝑀(𝑡,𝑦)
𝜕𝑦
dt
Method (ii)
Step 1: Knowing the equation N(t, y) =
𝜕Φ
,
𝜕𝑦
we determine Φ(t, y) up to arbitrary function of y along that
Φ(t, y) = ‫𝑡 𝑁 ׬‬, 𝑦 𝑑𝑦 + k(t).
Step 2: The k(t) is then determined by: k’(t) = M(t,y) - ‫׬‬
𝜕𝑁(𝑡,𝑦)
𝜕𝑡
dy.
Method (iii)
Step1: Since we have equations M(t,y) =
𝜕Φ
𝜕𝑡
and N(t,y) =
𝜕Φ
𝜕𝑦
, we simply integrate them
Step2: We do the comparison of h(y) and k(t) terms by inspection:
Φ(t, y) = ‫𝑡 𝑀 ׬‬, 𝑦 𝑑𝑡 + h(y),
Φ(t, y) = ‫𝑡 𝑁 ׬‬, 𝑦 𝑑𝑦 + k(t).
Example:
Find the general solution of the equation (3y + 𝑒 𝑡) + (3t + cos y)
𝑑𝑦
𝑑𝑡
=0.
How to solve ?
Step (i) we identify: M(t, y) = 3 y + 𝑒 𝑡 and N(t, y)= 3t + cos y ;
Step (ii) we compute and check
𝜕𝑀
𝜕𝑦
= 3 and
𝜕𝑁
=
𝜕𝑡
3, check if
If yes ! we deduce that the equation is exact.
Step (iii) we evaluate further using one of the 3 methods shown !
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
,
𝜕𝑡
METHOD 1
Example:
𝑑𝑦
Find the general solution of the equation (3y + 𝑒 𝑡) + (3t + cos y) 𝑑𝑡 = 0 .
(i) we have shown
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑡
(i) we have identify: M(t, y) = 3 y + 𝑒 𝑡 and N(t, y)= 3t + cos y ;
(ii) We evaluate using method 1
Since M(t,y) =
𝜕Φ
𝜕𝑡
, Integrate M(t,y) w.r.t ‘dt’ we obtained: Φ(t, y) = ‫𝑡 𝑀 ׬‬, 𝑦 𝑑𝑡 + h(y) = 3 y t + 𝑒 𝑡 + h(y).
Now differentiate Φ(t, y) by ‘y’ and identify it to N(t,y) =
So h’(y) = cos y which is then h(y) = sin y.
Hence the solution is Φ(t) = 3 yt + 𝑒 𝑡 + sin y.
𝜕Φ
𝜕𝑦
= 3t + h’(y) = 3t + cos y ;
Method 2
Example:
Find the general solution of the equation (3y + 𝑒 𝑡) + (3t + cos y)
(i) we have shown
𝜕𝑀
𝜕𝑦
=
𝑑𝑦
𝑑𝑡
=0.
𝜕𝑁
𝜕𝑡
(i) we have identify: M(t, y) = 3 y + 𝑒 𝑡 and N(t, y)= 3t + cos y ;
(ii) We evaluate using method 2
Since N(t,y) =
𝜕Φ
𝜕𝑦
, Integrate N(t,y) w.r.t ‘dy’ we obtained Φ(t, y) = ‫𝑡 𝑁 ׬‬, 𝑦 𝑑𝑦 + k(t) = 3 t y + sin y + k(t).
Now differentiate the above equation by ‘t’ gives M(t,y) =
𝜕Φ
𝜕𝑡
= 3y + k’(t) = 3y + 𝑒 𝑡 ;
So k’(t) = 𝑒 𝑡 which is then k(t) = 𝑒 𝑡 Hence the solution is Φ(t) = 3 y t + 𝑒 𝑡 + sin y.
Example:
Find the general solution of the equation (3y + 𝑒 𝑡) + (3t + cos y)
(i) we identify: M(t, y) = 3 y + 𝑒 𝑡 and N(t, y)= 3t + cos y ;
(ii) we have shown
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
,
𝜕𝑡
the equation is exact.
(iii) we evaluate using method 3
Method 3:
By comparing the two results:
Φ(t, y) = ‫𝑡 𝑀 ׬‬, 𝑦 𝑑𝑡 + h(y) = 3 y t + 𝑒 𝑡 + h(y) ;
Φ(t, y) = ‫𝑡 𝑁 ׬‬, 𝑦 𝑑𝑦 + k(t) = 3 t y + sin y + k(t)
We can deduce that h(y) = sin y and k(t) = 𝑒 𝑡.
Therefore just substitute either of it into the main expression,
Φ(t,y) = 3 y t + 𝑒 𝑡 + sin y.
𝑑𝑦
𝑑𝑡
=0.