1- First Order Ordinary Differential Equations
First of all we define the differential equations in general. A differential
equation is one that contains one or more derivative and may be classified
according to:
1- Type: Ordinary or Partial equation.
2- Order: Highest derivative in the equation.
3- Degree: The power on the highest order.
For example, take the equation;
(
𝑑3𝑦 2
)
𝑑𝑥 3
𝑑2𝑦 5
)
𝑑𝑥 2
+(
+
𝑦
1+ 𝑥 2
= ex
This equation is ordinary since it contains only one independent variable. It is third
order since the highest order is three. It is of second degree since the power on the
highest order is 2.
If there is more than one independent variable the equation is, then, partial
differential equation e.g.;
𝜕𝑦
𝜕𝑡
= 𝛼2
𝜕2 𝑦
𝜕𝑥 2
and it's solution is y = f(x,t)
The above equation is second order and first degree.
Note:
1- Differential equation of order n needs n number of integrations and it's
solution contains n number of constants.
2- The differential equation may be linear or non-linear depending on the
presence of the dependent variable y and it's derivatives in one term of the
equation, e.g.;
(i)
(ii)
(iii)
𝑑2𝑦
𝑑𝑥 2
𝑑2𝑦
𝑑𝑥 2
𝑑2𝑦
𝑑𝑥 2
𝑑𝑦
+ 4y 𝑑𝑥 +2y = 0
Non- linear equation.
𝑑𝑦
+ 4x 𝑑𝑥 +2y = 0
linear equation.
+ Sin y = 0
Non- linear equation since it contains
Sin y which is non- linear.
Classification of Ordinary Differential Equation of The First Order
1- Equations in which the variables are separable.
2- Homogeneous equation.
1
3- Linear equation.
4- Bernoulli's equation.
5- Exact equation.
1.1-
Equations in which the variables are separable
These type of equations can be reduced to the form,
A(x)dx + B(y)dy = 0
And, thus, can be integrated normally.
Example 1: Solve the equation,
y
𝑑𝑦
𝑑𝑥
+ x y2 = x
Solution:
y
𝑑𝑦
𝑑𝑥
+ ( y2 - 1) x = 0
𝑦
𝑦2− 1
dy + x dx = 0
1/2 ln ( y2- 1) +
𝑥2
2
+ C = 0;
where C is integration constant.
Example 2: Solve the equation,
Sin x Cos y dx + Cos x Sin y dy = 0
Solution:
Sin 𝑥
Cos 𝑥
dx +
Sin 𝑦
Cos 𝑦
dy = 0
-ln Cos x – ln Cos y + ln C = 0
ln
𝐶
Cos 𝑥 .Cos 𝑦
𝐶
Cos 𝑥 .Cos 𝑦
=0
=1
Cos 𝑥 . Cos 𝑦 = C
2
1.2-
Homogeneous equation: This type can be reduced to the form
A(x,y)dx + B(x,y)dy = 0
Where the functions A(x,y) and B(x,y) are of the same degree. And the solution starts
with the substitution,
y = v.x
𝑑𝑦
and
𝑑𝑣
=
𝑑𝑥
.𝑥 + 𝑣
𝑑𝑥
Where v is a dummy variable.
Example 3: Solve the equation,
(2 √𝑥. 𝑦 - x).dy + y.dx = 0
(1.1)
Solution:
We start with substitution,
y = v.x
∴
𝑑𝑦
𝑑𝑥
(1.2)
𝑑𝑣
=
𝑑𝑥
.𝑥+𝑣
(1.3)
Substitute (1.2) and (1.3) in (1.1),
(2.x √𝑣 - x). (x.
2x2 √𝑣 .
2x √𝑣 .
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
+ v) + v.x = 0, simplifying we get;
+ 2.x √𝑣 v – x2
𝑑𝑣
+ 2. √𝑣 v – x
𝑑𝑥
𝑑𝑣
𝑑𝑥
- v.x + v.x = 0
=0
(2x √𝑣 - 1).x.dv + 2. √𝑣 . v.dx = 0
2√𝑣− 1
dv
2𝑣√𝑣
Or,
𝑑𝑣
𝑣
+
- 1/2
𝑑𝑥
=0
𝑥
𝑑𝑣
𝑣 3/2
+
𝑑𝑥
𝑥
=0
∴ ln v – 1/2 (- v-1/2 ) +ln x – ln C = 0
𝑣.𝑥
𝐶
= 𝑒− 𝑣
−1/2
−1/2
= 𝑒 − (𝑦/𝑥)
1/2
∴ y = C 𝑒 − (𝑥/𝑦)
3
Example 4: Solve the equation,
(x3+y3) dy = x2y dx
(1.4)
Solution:
We start with substitution,
y = v.x
𝑑𝑦
∴
𝑑𝑥
𝑑𝑣
=
𝑑𝑥
.𝑥 +𝑣
Substitute for y and dy/dx in equation (1.4) to get;
𝑑𝑣
(x3 + v3x3).(v + x
(1 + v3).(v + x
v + v 4+ x
𝑑𝑣
𝑑𝑥
𝑑𝑣
)=v
𝑑𝑥
+v3 x
𝑑𝑣
v4+ (1+ v3) x
𝑑𝑥
) = x2.vx
𝑑𝑥
𝑑𝑣
𝑑𝑥
=v
=0
Dividing by x.v4 we get;
𝑑𝑥
𝑥
+
∴∫
1+ 𝑣 3
𝑑𝑥
𝑥
𝑣4
+∫
dv = 0
𝑑𝑣
𝑣4
+∫
𝑑𝑣
𝑣
=0
ln x + (- 1/3) v-3+ln v – ln C = 0
1 −3
x.v/C = 𝑒 3𝑣
1
(𝑦/𝑥)−3
1
(𝑥/𝑦)3
= 𝑒3
Finally, y = C 𝑒 3
1.3-
Linear equations
This type of equations has the general form,
𝑑𝑦
𝑑𝑥
+ P(x) y = Q(x)
And solved by an integrating factor (R), given by,
R = exp (∫ 𝑃(𝑥). 𝑑𝑥)
4
And the solution is,
R.y = ∫ 𝑅. 𝑄(𝑥). 𝑑𝑥 = C
Example 5: Solve the equation,
𝑑𝑦
x
𝑑𝑥
– y = x3
Solution:
𝑑𝑦
𝑑𝑥
-
𝑦
= x2
𝑥
−1
R = exp (∫
𝑥
. 𝑑𝑥) = e- ln x =
1
𝑥
R.y = ∫ 𝑅. 𝑄(𝑥). 𝑑𝑥 = C
1
1
𝑥
𝑥
∴ .y = ∫ . 𝑥 2 𝑑𝑥 + C
1
𝑥
1
𝑥
.y = ∫ 𝑥. 𝑑𝑥 + C
.y =
∴y=
𝑥2
2
𝑥3
2
+C
+ C.x
Example 6: Solve the equation,
𝑑𝑦
𝑑𝑥
+
R=𝑒
𝑥
1+ 𝑥
𝑥
y=
2
∫1+𝑥 𝑑𝑥
1+𝑥 (1+ 𝑥 2 )−1/2
1
𝑥 (1+ 𝑥 2 )2 + (1+ 𝑥 2 )
2 1/2
= 𝑒 ln(1+ 𝑥 ) = (1 + x2)1/2
∴ (1 + x2)1/2. y = ∫ (1 + x2)1/2
1+𝑥 (1+ 𝑥 2 )−1/2
[
1
𝑥 (1+ 𝑥 2 )2 + (1+ 𝑥 2 )
].dx + C
1
Taking (1 + 𝑥 2 )2 outside the denominator in RHS,
(1 + x2)1/2. y = ∫ (1 + x2)1/2
1
2 −1/2
1+𝑥 (1+ 𝑥 )
1 ].dx + C
1 [
2
(1+ 𝑥2 )2
1+𝑥 (1+ 𝑥 2 )−1/2
(1 + x2)1/2. y = ∫ [
1
𝑥 + (1+ 𝑥 )2
].dx + C
𝑥+ (1+ 𝑥 2 )2
5
1
(1 + x2)1/2. y = ln [𝑥 + (1 + 𝑥 2 )2 ] + C
1
∴ y=
1.4𝑑𝑦
𝑑𝑥
ln [𝑥+ (1+ 𝑥 2 )2 ]
(1+ x2 )1/2
+ (1+
𝐶
x2 )1/2
Bernoulli's equation
+ A(x).y = B(x).yn
First we solve the equation,
𝑑𝑦
𝑑𝑥
+ A(x).y = 0
Then we assume the solution as y = v.f(x), where v, here, is a dummy variable.
Example 7: Solve the equation,
𝑑𝑦
𝑑𝑥
- x.y = x3.y2
(1.5)
𝑑𝑦
First we solve,
𝑑𝑦
𝑦
𝑑𝑥
- x.y =0
- x.dx = 0
∴ ln y -
𝑥2
2
- lnC = 0
And, y = C 𝑒 𝑥
2 /2
Now, we put the solution in the form,
y = v. 𝑒 𝑥
And,
𝑑𝑦
𝑑𝑥
2 /2
=
𝑑𝑣
𝑑𝑥
(1.6)
𝑒𝑥
1
2 /2
+ 2xv (2. 𝑒 𝑥
2 /2
)
(1.7)
Substitute (1.6) and (1.7) in (1.5) to get,
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
𝑒𝑥
2 /2
+ x v 𝑒𝑥
2 /2
𝑒𝑥
2 /2
= x3.v2. 𝑒 𝑥
- x v 𝑒𝑥
2 /2
= x3.v2. 𝑒 𝑥
2
2
6
𝑑𝑣
∫ 𝑣2 = ∫ 𝑥3 𝑒 𝑥
2 /2
𝑑𝑥
Integrating by parts;
Let z = x2/2 ; then dz = x. dx
∴ ∫ 𝑥3 𝑒 𝑥
2 /2
∫ 𝑥 3 𝑒 𝑧 𝑑𝑧/𝑥 = ∫ 𝑥 2 𝑒 𝑧 𝑑𝑧 = ∫ 2. 𝑧. 𝑒 𝑧 𝑑𝑧 = 2 z. 𝑒 𝑧 2
2
2. 𝑒 𝑧 + K = x2. 𝑒 𝑥 /2 - 2. 𝑒 𝑥 /2 + K
𝑑𝑥
becomes
−1
Then we get,
𝑣
= x2. 𝑒 𝑥
2 /2
- 2. 𝑒 𝑥
2 /2
+K
Where K is the constant of integration. Substituting for v, we finally get;
y=
2
−𝑒 𝑥 /2
2
2
𝑥2.𝑒 𝑥 /2 − 2.𝑒 𝑥 /2+ 𝐾
Example 8: Solve the equation,
x
𝑑𝑦
𝑑𝑥
+ y = y2 ln x
First we solve,
∴
𝑑𝑦
𝑦
+
𝑑𝑥
𝑥
x
𝑑𝑦
𝑑𝑥
+y=0
=0
ln y + ln x – ln C = 0
y.x/C = 1
y=
𝐶
𝑥
Now put,
y=
𝑑𝑦
And,
𝑑𝑥
Substitute for y and
x(
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
𝑑𝑣 1
𝑑𝑥 𝑥
-
=
𝑣
𝑥
-v
+
𝑣2
𝑥2
𝑣
𝑥
1
𝑣
𝑥
𝑥
)+
2
=
𝑣2
𝑥2
𝑣
𝑥
=
𝑑𝑣 1
𝑑𝑥 𝑥
𝑑𝑦
𝑑𝑥
=
𝑣2
𝑥2
-v
1
𝑥2
in the original equation,
ln x
ln x
ln x
7
𝑑𝑣
∴∫
𝑣2
=∫
ln 𝑥
𝑥2
dx
( u.dv = u.v – ∫ 𝑣. 𝑑𝑢) ; putting u=lnx and dv=
Integrating RHS by parts,
1
𝑑𝑥
𝑣
𝑥
- = ln x. (-x-1) - ∫(−𝑥 −1 ).
1
ln 𝑥
𝑣
𝑥
- =-
𝑑𝑥
𝑥2
1
- +K
𝑥
Substitute for v,
−1
𝑥𝑦
=-
ln 𝑥
𝑥
𝑥
Finally, y =
1.5-
1
- +K
1
ln 𝑥−𝐾𝑥+1
The Exact equation
This has the general form,
A(x,y) dx + B(x,y) dy =0, on condition that;
𝜕𝐴(𝑥,𝑦)
𝜕𝑦
=
𝜕𝐵(𝑥,𝑦)
𝜕𝑥
Method of solution:
First we assume the solution is ∅(𝑥, 𝑦) = constant
Also, A =
𝑑∅
(1.8)
𝑑𝑥
And
B=
𝑑∅
(1.9)
𝑑𝑦
Integrating (1.8),
∅ = ∫A.dx
Differentiate with respect to y,
𝑑∅
𝑑𝑦
=
𝑑
𝑑𝑦
∫A.dx
(1.10)
Finally equating equations (1.9) and (1.10).
Example 9: Solve the equation,
8
(x3 – 3x2y + 2xy2).dx – (x3- 2x2y + y3).dy = 0
A(x,y)
B(x,y)
First we must check if the equation is exact,
𝜕𝐴
(
)x = - 3x2 + 4xy
𝜕𝑦
𝜕𝐵
(
𝜕𝑥
∴
)y = - 3x2 + 4xy
𝜕𝐴
𝜕𝑦
=
𝜕𝐵
𝜕𝑥
the equation is exact.
∅ = ∫A.dx = ∫(x3 – 3x2y + 2xy2)..dx
∴ ∅ = x4/4 – x3y + x2y2 + C(y)
(1.11)
Where C(y) is a constant that may be a function of y.
Differentiate (1.11) with respect to y,
𝑑∅
𝑑𝑦
= -x3 + 2x2y +
𝜕𝐶(𝑦)
(1.12)
𝜕𝑦
By definition equation (1.12) = B
∴
-x3 + 2x2y +
𝜕𝐶(𝑦)
𝜕𝑦
𝜕𝐶(𝑦)
𝜕𝑦
= – (x3- 2x2y + y3)
= - y3
∴ C(y) = - y4/4 - D ;
D is a constant of integration.
Substitute for C(y) in equation (1.11) to obtain the final solution,
∴ ∅ = x4/4 – x3y + x2y2- y4/4 = D
∴ ∅ = x4/4 – x3y + x2y2- y4/4
Example 10: Solve the equation,
Sin x.dy + y.Cos x.dx = 0
B
𝜕𝐴
𝜕𝑦
= Cos x
A
=
𝜕𝐵
𝜕𝑥
= Cos x ∴ exact.
∅ = ∫A.dx = ∫ y.Cos x.dx
9
∴ ∅ = y Sin x + C(y)
(1.13)
Differentiate with respect to y,
𝑑∅
𝑑𝑦
= Sin x +
𝜕𝐶(𝑦)
𝜕𝑦
∴ Sin x = Sin x +
𝜕𝐶(𝑦)
𝜕𝑦
; and this equation = B
𝜕𝐶(𝑦)
𝜕𝑦
=0
∴ C(y) = D (constant); substitute in equation (1.13);
∴ ∅ = y Sin x = D
∴ ∅ = y Sin x
Exercises
1- y 2 dx  x 2 dy  2 xydy
2- 4 x 2  6 xy  y 2 dx  x4 x  y dy  0
.3-
dy x  y cos x

dx
sin x  y
4- 3x  2 y 2 dx  2 xydy  0
5-
dy 4 y

x
dx x
6- x
7-
dy
 3y  x2
dx
x4

dy
 y y  x3
dx

8- ( x + x y 2 ) d x + ( y + x 2 y ) d y = 0
9- x 2 (
10- x(
dy 2
)  y2  0
dx
dy 2
dy
)  2 x  3 y   6 y  0
dx
dx
10