University of Science and Technology of Southern Philippines
College of Engineering and Architecture
Electrical Engineering Department
ES 208 DIFFERENTIAL EQUATIONS
Julius A. Nacional Jr.
Instructor - 1a
𝑑𝑦
𝑦′ =
𝑑𝑥
Prime Notation
𝑑2𝑦
𝑑𝑦
+
5
𝑑𝑥 2
𝑑𝑥
𝑥3
2𝑦
𝑑
𝑦 ′′ = 2
𝑑𝑥
− 4𝑦 = 𝑒 𝑥
𝑑2𝑦
𝑑𝑦
1−𝑥
−
4𝑥
+ 5𝑦 = cos 𝑥
𝑑𝑥 2
𝑑𝑥
𝑥5
2
𝑑𝑦
+
𝑑𝑥
4
+𝑦=0
𝑑4𝑦
𝑑2𝑦
3
− 2𝑥
+ 6𝑦 = 0
𝑑𝑥 4
𝑑𝑥 2
𝑑2𝑦
=
𝑑𝑥 2
𝑛
𝑦
𝑑𝑛 𝑦
= 𝑛
𝑑𝑥
3
𝑑3𝑦
𝑑𝑦
2𝑥
+
𝑥
−
5𝑦
=
3𝑒
𝑑𝑥 3
𝑑𝑥
𝑑3𝑦
𝑥
𝑑𝑥 3
𝑦 ′′′
𝑑3𝑦
= 3
𝑑𝑥
𝑑𝑦
1+
𝑑𝑥
𝑦 ′′ + 5 𝑦′
3
− 4𝑦 = 𝑒 𝑥
𝑥 3 𝑦′′′ + 𝑥𝑦′ − 5𝑦 = 3𝑒 2𝑥
1 − 𝑥 𝑦′′ − 4𝑥𝑦′ + 5𝑦 = cos 𝑥
𝑥 𝑦′′′
𝑥 5𝑦
4
2
+ 𝑦′
4
+𝑦=0
− 2𝑥 3 𝑦′′ + 6𝑦 = 0
4
𝑦′′ =
1 + 𝑦′
4
𝑑𝑦
𝑦=
𝑑𝑡
Dot Notation
𝑑2𝑦
𝑑𝑦
+
5
𝑑𝑡 2
𝑑𝑡
𝑑2𝑥
𝑥= 2
𝑑𝑡
3
− 4𝑦 = 𝑒 𝑡
3
𝑑
𝑥
𝑑𝑥
𝑡3 3 + 𝑡
− 5𝑥 = 3𝑒 2𝑡
𝑑𝑡
𝑑𝑡
𝑑2 𝑠
𝑑𝑠
1−𝑡
−
4𝑡
+ 5𝑠 = cos 𝜔𝑡
𝑑𝑡 2
𝑑𝑡
𝑑3𝑢
𝑡
𝑑𝑡 3
𝑑2𝑣
=
𝑑𝑣 2
2
𝑑3𝑠
𝑠= 3
𝑑𝑡
𝑑𝑢
+
𝑑𝑡
𝑑𝑣
1+
𝑑𝑡
𝑦+5 𝑦
3
− 4𝑦 = 𝑒 𝑡
𝑡 3 𝑥 + 𝑡 𝑥 − 5𝑥 = 3𝑒 2𝑡
1 − 𝑡 𝑠 − 4𝑡𝑠 + 5𝑠 = cos 𝜔𝑡
4
+𝑢 =0
𝑥 𝑢
2
+ 𝑢
4
+𝑢 =0
4
𝑣=
1+ 𝑣
4
Subscript Notation (partial derivatives)
𝜕2𝑢 𝜕2𝑢
+
=0
𝜕𝑥 2 𝜕𝑦 2
𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0
𝜕2𝑢 𝜕2𝑢
𝜕𝑢
=
−2
𝜕𝑥 2 𝜕𝑡 2
𝜕𝑡
𝑢𝑥𝑥 = 𝑢𝑡𝑡 − 2𝑢𝑡
𝜕𝑢
𝜕𝑣
=−
𝜕𝑦
𝜕𝑥
𝑢𝑦 = −𝑣𝑥
Classification by Linearity
An nth order Ordinary Differential Equation (ODE) is linear when
𝑑𝑛 𝑦
𝑑 𝑛−1 𝑦
𝑑𝑦
𝑎𝑛 𝑥
+
𝑎
𝑥
+
⋯
+
𝑎
𝑥
+ 𝑎0 𝑥 𝑦 = 𝑔 𝑥
𝑛−1
1
𝑑𝑥 𝑛
𝑑𝑥 𝑛−1
𝑑𝑥
𝑎𝑛 𝑥 𝑦
𝑛
+ 𝑎𝑛−1 𝑥 𝑦
𝑛−1
+ ⋯ + 𝑎1 𝑥 𝑦′ + 𝑎0 𝑥 𝑦 = 𝑔 𝑥
where
 The dependent variable 𝑦 and all its derivatives 𝑦 ′ , 𝑦 ′′ , … , 𝑦 𝑛 are
of the first degree, the power of each term involving 𝑦 is 1.
 The coefficients 𝑎0 , 𝑎1 , … , 𝑎𝑛 of 𝑦 ′ , 𝑦 ′′ , … , 𝑦 𝑛 depend at most on
the independent variable x.
A non-linear ODE is simply one that is not linear.
𝑑2𝑦
𝑑𝑦
+
5
𝑑𝑥 2
𝑑𝑥
3
− 4𝑦 = 𝑒 𝑥
𝑑3𝑦
𝑑𝑦
2𝑥
𝑥
+
𝑥
−
5𝑦
=
3𝑒
𝑑𝑥 3
𝑑𝑥
Non-linear
3
Linear
𝑑2𝑦
𝑑𝑦
1−𝑥
−
4𝑥
+ 5𝑦 = cos 𝑥
𝑑𝑥 2
𝑑𝑥
Linear
1 − 𝑦 𝑦 ′ + 2𝑦 = 𝑒 𝑥
Non-linear
𝑡5
𝑑4𝑦
𝑑2𝑦
3
− 2𝑡
+ 6𝑦 = 0
𝑑𝑡 4
𝑑𝑡 2
𝑑2𝑦
+ sin 𝑦 = 0
𝑑𝑥 2
Linear
Non-linear
𝑑𝑛 𝑦
𝑑 𝑛−1 𝑦
𝑑𝑦
𝑎𝑛 𝑥
+
𝑎
𝑥
+
⋯
+
𝑎
𝑥
+ 𝑎0 𝑥 𝑦 = 𝑔 𝑥
𝑛−1
1
𝑑𝑥 𝑛
𝑑𝑥 𝑛−1
𝑑𝑥
𝑛=1
𝑑𝑦
𝑎1 𝑥
+ 𝑎0 𝑥 𝑦 = 𝑔 𝑥
𝑑𝑥
𝑛=2
𝑑2𝑦
𝑑𝑦
𝑎2 𝑥
+ 𝑎1 𝑥
+ 𝑎0 𝑥 𝑦 = 𝑔 𝑥
𝑑𝑥 2
𝑑𝑥
Solution of Ordinary Differential Equation
Any function φ, defined on an interval I and possessing at least n derivatives
that are continuous in I, which when substituted into an nth-order ordinary
differential equation reduces the equation to an identity, is said to be a
solution of the equation on the interval.
The solution of an nth-order ODE
𝐹 𝑥, 𝑦, 𝑦 ′ , … , 𝑦
𝑛
=0
Is a function φ that possesses at least n derivatives and for which
𝐹 𝑥, 𝜑 𝑥 , 𝜑′ 𝑥 , … , 𝜑
𝑛
𝑥
=0
for all x in I.
Example
Verify that the indicated function is a solution of the given differential
equation on the interval −∞, ∞ .
1
𝑑𝑦
1 4
a. 𝑑𝑥 = 𝑥𝑦 2 ; 𝑦 = 16
𝑥
b. 𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0 ; 𝑦 = 𝑥𝑒 𝑥
Solution:
a)
𝑑𝑦
1
= 16
𝑑𝑥
1 4
𝑦 = 16
𝑥
4𝑥 3
= 14𝑥 3
Substituting to the differential equation
𝑑𝑦
1
= 𝑥𝑦 2
𝑑𝑥
1 3
4𝑥
=𝑥
1 4
16𝑥
1
2
=𝑥
1 2
4𝑥
1 3
= 𝑥
4
b)
𝑦 = 𝑥𝑒 𝑥
𝑦 ′ = 𝑥𝑒 𝑥 + 𝑒 𝑥
𝑦 ′′ = 𝑥𝑒 𝑥 + 𝑒 𝑥 + 𝑒 𝑥 = 𝑥𝑒 𝑥 + 2𝑒 𝑥
Substituting to the differential equation
𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0
𝑥𝑒 𝑥 + 2𝑒 𝑥 − 2 𝑥𝑒 𝑥 + 𝑒 𝑥 + 𝑥𝑒 𝑥 = 0
0=0
Example
Verify that the indicated family of function 𝑦 = 𝑐1 𝑥 −1 + 𝑐2 𝑥 + 𝑐3 𝑥 ln 𝑥 + 4𝑥 2
3
𝑑 𝑦
equation 𝑥 3 𝑑𝑥 3
2
𝑑 𝑦
2𝑥 2 𝑑𝑥 2
𝑑𝑦
is a solution of the differential
+
− 𝑥 𝑑𝑥 + 𝑦 = 12𝑥 2 .
Assume an appropriate interval I of definition and 𝑐1 , 𝑐2 and 𝑐3 are arbitrary
constants or parameters.
Solution:
𝑦 = 𝑐1 𝑥 −1 + 𝑐2 𝑥 + 𝑐3 𝑥 ln 𝑥 + 4𝑥 2
𝑑𝑦
1
−2
= 𝑐1 −𝑥
+ 𝑐2 + 𝑐3 𝑥
+ ln 𝑥 + 8𝑥 = −𝑐1 𝑥 −2 + 𝑐2 + 𝑐3 +𝑐3 ln 𝑥 +8x
𝑑𝑥
𝑥
𝑑2𝑦
−3 +𝑐 𝑥 −1 +8
=
2𝑐
𝑥
1
3
𝑑𝑥 2
𝑑3𝑦
= −6𝑐1 𝑥 −4 − 𝑐3 𝑥 −2
3
𝑑𝑥
Substituting to the differential equation
𝑥3
𝑑3𝑦
𝑑2𝑦
𝑑𝑦
2
+ 2𝑥
−𝑥
+ 𝑦 = 12𝑥 2
3
2
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑3𝑦
3
−4
−2
−1
𝑥
=
𝑥
−6𝑐
𝑥
−
𝑐
𝑥
=
−6𝑐
𝑥
− 𝑐3 𝑥
1
3
1
𝑑𝑥 3
3
𝑑2𝑦
2
−3
−1
−1
2
2𝑥
=
2𝑥
2𝑐
𝑥
+𝑐
𝑥
+8
=
4𝑐
𝑥
+2𝑐
𝑥+16𝑥
1
3
1
3
𝑑𝑥 2
2
𝑑𝑦
−𝑥
= −𝑥 −𝑐1 𝑥 −2 + 𝑐2 + 𝑐3 +𝑐3 ln 𝑥 +8x = 𝑐1 𝑥 −1 − 𝑐2 𝑥 − 𝑐3 𝑥 − 𝑐3 𝑥 ln 𝑥 −8𝑥 2
𝑑𝑥
𝑦 = 𝑐1 𝑥 −1 + 𝑐2 𝑥 + 𝑐3 𝑥 ln 𝑥 + 4𝑥 2
12𝑥 2 = 12𝑥 2
Example
Verify that the indicated pair of functions
𝑥 = cos 2𝑡 + sin 2𝑡 + 15𝑒 𝑡
𝑦 = − cos 2𝑡 − sin 2𝑡 − 15𝑒 𝑡
is a solution of the system of differential equations
𝑑2 𝑥
𝑑𝑡 2
𝑑2 𝑦
𝑑𝑡 2
= 4𝑦 + 𝑒 𝑡
= 4𝑥 − 𝑒 𝑡
on the interval −∞, ∞ .
Solution:
𝑥 = cos 2𝑡 + sin 2𝑡 + 15𝑒 𝑡
𝑦 = −cos 2𝑡 − sin 2𝑡 − 15𝑒 𝑡
𝑑𝑥
= −2 sin 2𝑡 + 2 cos 2𝑡 + 15𝑒 𝑡
𝑑𝑡
𝑑𝑦
= 2 sin 2𝑡 − 2 cos 2𝑡 − 15𝑒 𝑡
𝑑𝑡
𝑑2𝑥
= −4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡
2
𝑑𝑡
𝑑2𝑦
1 𝑡
=
4
cos
2𝑡
+
4
sin
2𝑡
−
5𝑒
𝑑𝑡 2
Substituting to the differential equation
𝑑2𝑥
= 4𝑦 + 𝑒 𝑡
2
𝑑𝑡
−4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡 = 4 −cos 2𝑡 − sin 2𝑡 − 15𝑒 𝑡 + 𝑒 𝑡
= −4cos 2𝑡 − 4 sin 2𝑡 − 45𝑒 𝑡 + 𝑒 𝑡
−4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡 = −4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡
𝑑2𝑦
= 4𝑥 − 𝑒 𝑡
2
𝑑𝑡
4 cos 2𝑡 + 4 sin 2𝑡 − 15𝑒 𝑡 = 4 cos 2𝑡 + sin 2𝑡 + 15𝑒 𝑡 − 𝑒 𝑡
= 4 cos 2𝑡 + 4 sin 2𝑡 + 45𝑒 𝑡 − 𝑒 𝑡
4 cos 2𝑡 + 4 sin 2𝑡 − 15𝑒 𝑡 = 4 cos 2𝑡 + 4 sin 2𝑡 − 15𝑒 𝑡
Example
Find the values of m so that the function
𝑦 = 𝑒 𝑚𝑥
is a solution of the differential equation
2𝑦 ′′ + 7𝑦 ′ − 4𝑦 = 0
Solution:
𝑦 = 𝑒 𝑚𝑥
𝑦′ = 𝑚𝑒 𝑚𝑥
𝑦′′ = 𝑚2 𝑒 𝑚𝑥
Substituting to the differential equation
2𝑦 ′′ + 7𝑦 ′ − 4𝑦 = 0
2𝑚2 𝑒 𝑚𝑥 + 7𝑚𝑒 𝑚𝑥 − 4𝑒 𝑚𝑥 = 0
2𝑚2 + 7𝑚 − 4 𝑒 𝑚𝑥 = 0
2𝑚2 + 7𝑚 − 4 = 0
2𝑚 − 1 𝑚 + 4 = 0
2𝑚 − 1 = 0
𝑚=1 2
𝑚+4=0
𝑚 = −4
Assignment #1
Exercises 1-19 of ref ii
Problems 1-13,15-20 of ref iv
Solution of Ordinary Differential Equation (ref I,pp 6)
Any function φ, defined on an interval I and possessing at least n derivatives
that are continuous in I, which when substituted into an nth-order ordinary
differential equation reduces the equation to an identity, is said to be a
solution of the equation on the interval.
The solution of an nth-order ODE
𝐹 𝑥, 𝑦, 𝑦 ′ , … , 𝑦
𝑛
=0
Is a function φ that possesses at least n derivatives and for which
𝐹 𝑥, 𝜑 𝑥 , 𝜑′ 𝑥 , … , 𝜑
𝑛
𝑥
=0
for all x in I.
Example (ref I, pp 7
Verify that the indicated function is a solution of the given differential
equation on the interval −∞, ∞ .
1
𝑑𝑦
1 4
a. 𝑑𝑥 = 𝑥𝑦 2 ; 𝑦 = 16
𝑥
Solution:
a)
𝑑𝑦
1
= 16
𝑑𝑥
1 4
𝑦 = 16
𝑥
4𝑥 3
= 14𝑥 3
Substituting to the differential equation
𝑑𝑦
1
= 𝑥𝑦 2
𝑑𝑥
1 3
4𝑥
=𝑥
1 4
16𝑥
1
2
=𝑥
1 2
4𝑥
1 3
= 𝑥
4
b)
𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0 ; 𝑦 = 𝑥𝑒 𝑥
𝑦 = 𝑥𝑒 𝑥
𝑦 ′ = 𝑥𝑒 𝑥 + 𝑒 𝑥
𝑦 ′′ = 𝑥𝑒 𝑥 + 𝑒 𝑥 + 𝑒 𝑥 = 𝑥𝑒 𝑥 + 2𝑒 𝑥
Substituting to the differential equation
𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0
𝑥𝑒 𝑥 + 2𝑒 𝑥 − 2 𝑥𝑒 𝑥 + 𝑒 𝑥 + 𝑥𝑒 𝑥 = 0
0=0
Example (ref i
Verify that the indicated family of function 𝑦 = 𝑐1 𝑥 −1 + 𝑐2 𝑥 + 𝑐3 𝑥 ln 𝑥 + 4𝑥 2
3
𝑑 𝑦
equation 𝑥 3 𝑑𝑥 3
2
𝑑 𝑦
2𝑥 2 𝑑𝑥 2
𝑑𝑦
is a solution of the differential
+
− 𝑥 𝑑𝑥 + 𝑦 = 12𝑥 2 .
Assume an appropriate interval I of definition and 𝑐1 , 𝑐2 and 𝑐3 are arbitrary
constants or parameters.
Solution:
𝑦 = 𝑐1 𝑥 −1 + 𝑐2 𝑥 + 𝑐3 𝑥 ln 𝑥 + 4𝑥 2
𝑑𝑦
1
−2
= 𝑐1 −𝑥
+ 𝑐2 + 𝑐3 𝑥
+ ln 𝑥 + 8𝑥 = −𝑐1 𝑥 −2 + 𝑐2 + 𝑐3 +𝑐3 ln 𝑥 +8x
𝑑𝑥
𝑥
𝑑2𝑦
−3 +𝑐 𝑥 −1 +8
=
2𝑐
𝑥
1
3
𝑑𝑥 2
𝑑3𝑦
= −6𝑐1 𝑥 −4 − 𝑐3 𝑥 −2
3
𝑑𝑥
Substituting to the differential equation
𝑥3
𝑑3𝑦
𝑑2𝑦
𝑑𝑦
2
+ 2𝑥
−𝑥
+ 𝑦 = 12𝑥 2
3
2
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑3𝑦
3
−4
−2
−1
𝑥
=
𝑥
−6𝑐
𝑥
−
𝑐
𝑥
=
−6𝑐
𝑥
− 𝑐3 𝑥
1
3
1
𝑑𝑥 3
3
𝑑2𝑦
2
−3
−1
−1
2
2𝑥
=
2𝑥
2𝑐
𝑥
+𝑐
𝑥
+8
=
4𝑐
𝑥
+2𝑐
𝑥+16𝑥
1
3
1
3
𝑑𝑥 2
2
𝑑𝑦
−𝑥
= −𝑥 −𝑐1 𝑥 −2 + 𝑐2 + 𝑐3 +𝑐3 ln 𝑥 +8x = 𝑐1 𝑥 −1 − 𝑐2 𝑥 − 𝑐3 𝑥 − 𝑐3 𝑥 ln 𝑥 −8𝑥 2
𝑑𝑥
𝑦 = 𝑐1 𝑥 −1 + 𝑐2 𝑥 + 𝑐3 𝑥 ln 𝑥 + 4𝑥 2
12𝑥 2 = 12𝑥 2
Example
Verify that the indicated pair of functions
𝑥 = cos 2𝑡 + sin 2𝑡 + 15𝑒 𝑡
𝑦 = − cos 2𝑡 − sin 2𝑡 − 15𝑒 𝑡
is a solution of the system of differential equations
𝑑2 𝑥
𝑑𝑡 2
𝑑2 𝑦
𝑑𝑡 2
= 4𝑦 + 𝑒 𝑡
= 4𝑥 − 𝑒 𝑡
on the interval −∞, ∞ .
Solution:
𝑥 = cos 2𝑡 + sin 2𝑡 + 15𝑒 𝑡
𝑦 = −cos 2𝑡 − sin 2𝑡 − 15𝑒 𝑡
𝑑𝑥
= −2 sin 2𝑡 + 2 cos 2𝑡 + 15𝑒 𝑡
𝑑𝑡
𝑑𝑦
= 2 sin 2𝑡 − 2 cos 2𝑡 − 15𝑒 𝑡
𝑑𝑡
𝑑2𝑥
= −4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡
2
𝑑𝑡
𝑑2𝑦
1 𝑡
=
4
cos
2𝑡
+
4
sin
2𝑡
−
5𝑒
𝑑𝑡 2
Substituting to the differential equation
𝑑2𝑥
= 4𝑦 + 𝑒 𝑡
2
𝑑𝑡
−4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡 = 4 −cos 2𝑡 − sin 2𝑡 − 15𝑒 𝑡 + 𝑒 𝑡
= −4cos 2𝑡 − 4 sin 2𝑡 − 45𝑒 𝑡 + 𝑒 𝑡
−4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡 = −4 cos 2𝑡 − 4 sin 2𝑡 + 15𝑒 𝑡
𝑑2𝑦
= 4𝑥 − 𝑒 𝑡
2
𝑑𝑡
4 cos 2𝑡 + 4 sin 2𝑡 − 15𝑒 𝑡 = 4 cos 2𝑡 + sin 2𝑡 + 15𝑒 𝑡 − 𝑒 𝑡
= 4 cos 2𝑡 + 4 sin 2𝑡 + 45𝑒 𝑡 − 𝑒 𝑡
4 cos 2𝑡 + 4 sin 2𝑡 − 15𝑒 𝑡 = 4 cos 2𝑡 + 4 sin 2𝑡 − 15𝑒 𝑡
Example
Find the values of m so that the function
𝑦 = 𝑒 𝑚𝑥
is a solution of the differential equation
2𝑦 ′′ + 7𝑦 ′ − 4𝑦 = 0
Solution:
𝑦 = 𝑒 𝑚𝑥
𝑦′ = 𝑚𝑒 𝑚𝑥
𝑦′′ = 𝑚2 𝑒 𝑚𝑥
Substituting to the differential equation
2𝑦 ′′ + 7𝑦 ′ − 4𝑦 = 0
2𝑚2 𝑒 𝑚𝑥 + 7𝑚𝑒 𝑚𝑥 − 4𝑒 𝑚𝑥 = 0
2𝑚2 + 7𝑚 − 4 𝑒 𝑚𝑥 = 0
2𝑚2 + 7𝑚 − 4 = 0
2𝑚 − 1 𝑚 + 4 = 0
2𝑚 − 1 = 0
𝑚=1 2
𝑚+4=0
𝑚 = −4
Assignment #1
Problems 1-13,15-28 of ref iv, pp 24-25
Problem Set #1
#s 1-29 of exercise 1.3 of ref i, pp 30-33
Deadline Sept. 21, 2022
𝑓𝐷 = 𝛾𝑣
𝑤 = 𝑚𝑔
𝐹 = 𝑤 − 𝑓𝐷 = 𝑚𝑔 − 𝛾𝑣
𝐹 = 𝑚𝑎
𝑚𝑎 = 𝑚𝑔 − 𝛾𝑣
𝑎=
𝑚
𝑑𝑣
𝑑𝑡
𝑑𝑣
= 𝑚𝑔 − 𝛾𝑣
𝑑𝑡
𝑑𝑣
𝛾
=𝑔− 𝑣
𝑑𝑡
𝑚
𝑔 = 9.8 𝑚
𝑠2
𝑑𝑣
𝛾
= 9.8 − 𝑣
𝑑𝑡
𝑚
𝑚 = 10 𝑘𝑔
𝛾=2
𝑑𝑣
𝑣
= 9.8 −
𝑑𝑡
5
𝑘𝑔
𝑠
𝑣 𝑡 =?
A first-order differential equation of the form
𝑑𝑦
=𝑔 𝑥 ℎ 𝑦
𝑑𝑥
is said to be separable or to have separable variables.
𝑝 𝑥, 𝑦 𝑑𝑥 + 𝑞 𝑥, 𝑦 𝑑𝑦 = 0
r 𝑥 𝑑𝑥 + 𝑠 𝑦 𝑑𝑦 = 0
Example 1 ( ref ii pp 18)
Solution
𝑑𝑦
𝑑𝑥
=2
𝑦
𝑥
𝑑𝑦
𝑑𝑥
= 2
𝑦
𝑥
ln 𝑦 = 2 ln 𝑥 + 𝑐1
𝑎 ln 𝑏 = ln 𝑏𝑎
ln 𝑦 = ln 𝑥 2 + 𝑐1
𝑒 ln 𝑦 = 𝑒 ln 𝑥
2 +𝑐
1
2
𝑒 ln 𝑦 = 𝑒 ln 𝑥 𝑒 𝑐1
𝑒 ln 𝑎 = 𝑎
𝑦 = 𝑐𝑥 2
𝑒 𝑐1 = 𝑐
General Solution
Example 2 ( ref ii pp 20)
Solution
𝑑𝑥
𝑑𝑦
+
=0
1 + 𝑥2 1 + 𝑦2
𝑑𝑢
1
𝑢
−1
=
tan
+𝑐
𝑎2 + 𝑢2 𝑎
𝑎
tan−1 𝑥 + tan−1 𝑦 = 𝑐
General Solution
@ 𝑥 = 0, 𝑦 = −1
1
0− 𝜋 =𝑐
4
1
𝑐=− 𝜋
4
1
tan−1 𝑥 + tan−1 𝑦 = − 𝜋
4
Particular Solution
Example 3 ( ref ii pp 22)
Solution
𝑦 𝑑𝑦
2𝑥 𝑑𝑥 −
=0
𝑦+1
𝑦
1
=1−
𝑦+1
𝑦+1
2𝑥 𝑑𝑥 − 1 −
2𝑥 𝑑𝑥 −
1
𝑑𝑦 = 0
𝑦+1
1
1−
𝑑𝑦 = 𝑐
𝑦+1
𝑥 2 − 𝑦 + ln 𝑦 + 1 = 𝑐
@ 𝑥 = 0, 𝑦 = −2
2 + ln −1 = 𝑐
𝑐=2
𝑥 2 − 𝑦 + ln 𝑦 + 1 = 2
Example 4 ( ref iv pp 42)
Solution
1 − 𝑦 2 𝑑𝑦 = 𝑥 2 𝑑𝑥
1 3 1 3 𝑐
𝑦− 𝑦 = 𝑥 +
3
3
3
3𝑦 − 𝑦 3 = 𝑥 3 + 𝑐
Example 5 ( ref iv pp 45)
Solution
2 𝑦 − 1 𝑑𝑦 = 3𝑥 2 + 4𝑥 + 2 𝑑𝑥
𝑦 2 − 2𝑦 = 𝑥 3 + 2𝑥 2 + 2𝑥 + 𝑐
@ 𝑥 = 0, 𝑦 = −1
1+2=𝑐
𝑦−1
= 𝑥 3 + 2𝑥 2 + 2𝑥 + 𝑐1
𝑦 2 − 2𝑦 + 1 = 𝑥 3 + 2𝑥 2 + 2𝑥 + 𝑐1
@ 𝑥 = 0, 𝑦 = −1
𝑐=3
𝑦 2 − 2𝑦 = 𝑥 3 + 2𝑥 2 + 2𝑥 + 3
2
1 + 2 + 1 = 𝑐1
𝑐1 = 4
𝑦 2 − 2𝑦 + 1 = 𝑥 3 + 2𝑥 2 + 2𝑥 + 4
𝑦 2 − 2𝑦 = 𝑥 3 + 2𝑥 2 + 2𝑥 + 3
Example 6 ( ref i pp 52)
Solve the differential equation
𝑑𝑦
2𝑦 + 3
=
𝑑𝑥
4𝑥 + 5
2
Solution
𝑑𝑦
2𝑦 + 3
=
𝑑𝑥
4𝑥 + 5
2𝑦 + 3
−
−2
2
2𝑦 + 3
=
4𝑥 + 5
𝑑𝑦 = 4𝑥 + 5
1
2𝑦 + 3
2
−1
=−
−2
2
2
𝑑𝑥
1
4𝑥 + 5
4
−1
−
𝑐
4
2 4𝑥 + 5 = 2𝑦 + 3 + 𝑐 2𝑦 + 3 4𝑥 + 5
8𝑥 − 2𝑦 + 7 = 𝑐 8𝑥𝑦 + 12𝑥 + 10𝑦 + 15
Example 7 ( ref i pp 52)
Solve the differential equation
sin 3𝑥 𝑑𝑥 + 2𝑦𝑐𝑜𝑠 3 3𝑥 𝑑𝑦 = 0
Solution
sin 3𝑥 𝑐𝑜𝑠 −3 3𝑥 𝑑𝑥 + 2𝑦 𝑑𝑦 = 0
1
𝑐
𝑐𝑜𝑠 −2 3𝑥 + 𝑦 2 =
6
6
𝑠𝑒𝑐 2 3𝑥 + 6𝑦 2 = 𝑐
Example 7 ( ref i pp 52)
Solve the differential equation
𝑑𝑦 𝑥𝑦 + 2𝑦 − 𝑥 − 2
=
𝑑𝑥 𝑥𝑦 − 3𝑦 + 𝑥 − 3
Solution
𝑥+2 𝑦−1
𝑑𝑦 𝑥𝑦 + 2𝑦 − 𝑥 − 2 𝑦 𝑥 + 2 − 𝑥 + 2
=
=
=
𝑥−3 𝑦+1
𝑑𝑥 𝑥𝑦 − 3𝑦 + 𝑥 − 3 𝑦 𝑥 − 3 + 𝑥 − 3
𝑦+1
𝑥+2
𝑑𝑦 =
𝑑𝑥
𝑦−1
𝑥−3
2
1+
𝑑𝑦 =
𝑦−1
5
1+
𝑑𝑥 + 𝑐
𝑥−3
𝑦 + 2 ln 𝑦 − 1 = 𝑥 + 5 ln 𝑥 − 3 + 𝑐
Example 8 ( ref i pp 52)
Find the particular solution of the given initial-value problem
𝑒 2𝑦 − 𝑦 cos 𝑥
𝑑𝑦
= 𝑒 𝑦 sin 2𝑥
𝑑𝑥
𝑦 0 =0
Solution
𝑒 2𝑦 − 𝑦
sin 2𝑥
𝑑𝑦
=
𝑑𝑥
𝑒𝑦
cos 𝑥
sin 2𝜃 = 2 sin 𝜃 cos 𝜃
𝑒 𝑦 − 𝑦𝑒 −𝑦 𝑑𝑦 =
𝑢 𝑑𝑣 = 𝑢𝑣 −
𝑣 𝑑𝑢
𝑢=𝑦
𝑑𝑣 = 𝑒 −𝑦 𝑑𝑦
𝑑𝑢 = 𝑑𝑦
𝑣 = −𝑒 −𝑦
2 sin 𝑥 𝑑𝑥 + 𝑐
𝑒 𝑦 − −𝑦𝑒 −𝑦 − 𝑒 −𝑦 = −2 cos 𝑥 + 𝑐
𝑒 𝑦 + 𝑦𝑒 −𝑦 + 𝑒 −𝑦 = −2 cos 𝑥 + 𝑐
@𝑥 = 0, 𝑦 = 0
1 + 0 + 1 = −2 + 𝑐
𝑐=4
𝑒 𝑦 + 𝑦𝑒 −𝑦 + 𝑒 −𝑦 = 4 − 2 cos 𝑥
𝑦𝑒 −𝑦 𝑑𝑦 = −𝑦𝑒 −𝑦 +
𝑒 −𝑦 𝑑𝑦
𝑦𝑒 −𝑦 𝑑𝑦 = −𝑦𝑒 −𝑦 − 𝑒 −𝑦
Example 9 ( ref i pp 52)
Find the particular solution of the given initial-value problem
𝑑𝑦 𝑦 2 − 1
=
,𝑦 2 = 2
𝑑𝑥 𝑥 2 − 1
Solution
𝑑𝑦
𝑑𝑥
=
𝑦2 − 1 𝑥2 − 1
1
1
1
1
−
𝑑𝑦 =
2
𝑦−1 𝑦+1
2
1
𝐴
𝐵
=
+
𝑢2 − 1 𝑢 + 1 𝑢 − 1
1
1
−
𝑑𝑥 + 𝑐1
𝑥−1 𝑥+1
1=𝐴 𝑢−1 +𝐵 𝑢+1
@𝑢 =1
@ 𝑢 = −1
1
1
ln 𝑦 − 1 − ln 𝑦 + 1 = ln 𝑥 − 1 − ln 𝑥 + 1 + 𝑐1
2
2
𝐵 = 12
𝐴 = −12
1
𝑦−1
1
𝑥−1
ln
= ln
+ 𝑐1
1
1
1
2
𝑦+1
2
𝑥+1
2
2
=
−
𝑢2 − 1 𝑢 − 1 𝑢 + 1
𝑦−1
𝑥−1
ln
− ln
= 2𝑐1
𝑦+1
𝑥+1
𝑎
ln 𝑎 − ln 𝑏 = ln
𝑦−1
𝑏
𝑦−1 𝑥+1
𝑥𝑦 − 𝑥 + 𝑦 − 1
𝑦+1
ln
= ln
= ln
= 2𝑐1
𝑥−1
𝑦+1 𝑥−1
𝑥𝑦 + 𝑥 − 𝑦 − 1
𝑥+1
𝑥𝑦 − 𝑥 + 𝑦 − 1
ln
= 2𝑐1
𝑥𝑦 + 𝑥 − 𝑦 − 1
𝑥𝑦 − 𝑥 + 𝑦 − 1
= 𝑒 2𝑐1 = 𝑐
𝑥𝑦 + 𝑥 − 𝑦 − 1
𝑥𝑦 − 𝑥 + 𝑦 − 1 = c 𝑥𝑦 + 𝑥 − 𝑦 − 1
@ 𝑥 = 2, 𝑦 = 2
4−2+2−1=c 4+2−2−1
𝑐=1
𝑥𝑦 − 𝑥 + 𝑦 − 1 = 𝑥𝑦 + 𝑥 − 𝑦 − 1
2𝑦 = 2𝑥
𝒚=𝒙
𝑒 ln 𝑎 = 𝑎
Assignment #2
Exercises 7-37 on pp. 23 of ref ii
Problems 1-28 on pp. 47- 49 of ref iv
𝑓 𝑥, 𝑦 = 𝑥 2 − 3𝑥𝑦 + 4𝑦 2
𝑓 𝜆𝑥, 𝜆𝑦 = 𝜆𝑥
2
− 3 𝜆𝑥 𝜆𝑦 + 4 𝜆𝑦
𝑓 𝜆𝑥, 𝜆𝑦 = 𝜆2 𝑥 2 − 3𝜆2 𝑥𝑦 + 4𝜆2 𝑦 2
𝑓 𝜆𝑥, 𝜆𝑦 = 𝜆2 𝑥 2 − 3𝑥𝑦 + 4𝑦 2
𝑓 𝜆𝑥, 𝜆𝑦 = 𝜆2 𝑓 𝑥, 𝑦
2
𝑥
M and N are homogeneous of the same degree
𝑥 = 𝑢𝑦
𝑜𝑟
𝑦
𝑦
−1 𝑒
𝑥
𝑦−𝑥 𝑒
𝑦
𝑥
𝑥
=𝑐
=𝑐
𝑦 = 𝑣𝑥
Example 1 ( ref ii pp. 26)
Solution:
𝑦 = 𝑣𝑥
𝑥 2 − 𝑥 𝑣𝑥 + 𝑣𝑥
𝑑𝑦 = 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣
2
𝑑𝑥 − 𝑥 𝑣𝑥 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
𝑥 2 1 − 𝑣 + 𝑣 2 𝑑𝑥 − 𝑥 2 𝑣 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
1 − 𝑣 + 𝑣 2 𝑑𝑥 − 𝑣 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
1 − 𝑣 𝑑𝑥 − 𝑣𝑥 𝑑𝑣 = 0
𝑑𝑥 𝑣 𝑑𝑣
−
=0
𝑥
1−𝑣
𝑑𝑥 𝑣 𝑑𝑣
+
=0
𝑥
𝑣−1
𝑑𝑥
1
+ 1+
𝑑𝑣 = 0
𝑥
𝑣−1
ln 𝑥 + 𝑣 + ln 𝑣 − 1 = 𝑐1
𝑒
ln 𝑥 +𝑣+ln 𝑣−1
𝑥 𝑣 − 1 𝑒𝑣 = 𝑐
= 𝑒 𝑐1
Example 1 ( ref ii pp. 26)
Solution:
𝑥 = 𝑢𝑦
𝑢𝑦
2
𝑑𝑥 = 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢
− 𝑢𝑦 𝑦 + 𝑦 2 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 − 𝑢𝑦 𝑦 𝑑𝑦 = 0
𝑦 2 𝑢2 − 𝑢 + 1 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 − 𝑦 2 𝑢 𝑑𝑦 = 0
𝑢2 − 𝑢 𝑢 𝑑𝑦 + 𝑦 𝑢2 − 𝑢 + 1 𝑑𝑢 = 0
𝑑𝑦
𝑢2 − 𝑢 + 1
+
𝑑𝑢 = 0
𝑦
𝑢2 𝑢 − 1
𝑑𝑦
1
1
+
− 2 𝑑𝑢 = 0
𝑦
𝑢−1 𝑢
1
ln 𝑦 + ln 𝑢 − 1 + = ln 𝑐
𝑢
𝑦 𝑢−1 𝑒
1
𝑢
𝑥
𝑦
𝑦
−1 𝑒
𝑦
= 𝑐1
𝑥
= 𝑐1
𝑢2 − 𝑢 + 1 𝐴 𝐵
𝐶
= + 2+
𝑢2 𝑢 − 1
𝑢 𝑢
𝑢−1
𝑢2 − 𝑢 + 1 = 𝐴 𝑢2 − 𝑢 + 𝐵 𝑢 − 1 + 𝐶𝑢2
𝐵 = −1
𝐵 − 𝐴 = −1
𝐴+𝐶 =1
𝑥−𝑦 𝑒
𝐴=0
𝐶=1
𝑦
𝑦−𝑥 𝑒
𝑦
𝑥
= 𝑐1
𝑥
=𝑐
Example 2 ( ref ii pp. 27)
Solution:
𝑥 = 𝑢𝑦
𝑑𝑥 = 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢
2
𝑦2
𝑢𝑦 𝑦 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 + 𝑢𝑦 +
𝑑𝑦 = 0
𝑢𝑦 2 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 + 𝑦 2 𝑢2 + 1 𝑑𝑦 = 0
𝑢 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 + 𝑢2 + 1 𝑑𝑦 = 0
𝑢𝑦 𝑑𝑢 + 2𝑢2 + 1 𝑑𝑦 = 0
𝑢 𝑑𝑢
𝑑𝑦
+
=0
2𝑢2 + 1 𝑦
1
1
ln 2𝑢2 + 1 + ln 𝑦 = ln 𝑐
4
4
ln 2𝑢2 + 1 + 4 ln 𝑦 = ln 𝑐
ln 2𝑢2 + 1 𝑦 4 = ln 𝑐
2𝑢2 + 1 𝑦 4 = 𝑐
𝑥
2
𝑦
2
+ 1 𝑦4 = 𝑐
2𝑥 2 + 𝑦 2 𝑦 2 = 𝑐
Example 2 ( ref ii pp. 27)
Solution:
𝑦 = 𝑣𝑥
𝑑𝑦 = 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣
𝑥 𝑣𝑥 𝑑𝑥 + 𝑥 2 + 𝑣𝑥
𝑣𝑥 2 𝑑𝑥
𝑥2
2
𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
𝑣2
+
1+
𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
𝑣𝑑𝑥 + 1 + 𝑣 2 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
2𝑣 + 𝑣 3 𝑑𝑥 + 1 + 𝑣 2 𝑥 𝑑𝑣 = 0
𝑑𝑥
1 + 𝑣2
+
𝑥
𝑣 2 + 𝑣2
1 + 𝑣2
𝐴
𝐵
2𝑣𝐶
=
+
+
𝑣 2 + 𝑣2
𝑣 2 + 𝑣2 2 + 𝑣2
𝑑𝑣 = 0
𝑑𝑥
1 1
1
2𝑣
+
+
𝑥
2 𝑣
4 2 + 𝑣2
𝑑𝑣 = 0
1
1
1
ln 𝑥 + ln 𝑣 + ln 2 + 𝑣 2 = ln 𝑐
2
4
4
4ln 𝑥 + 2 ln 𝑣 + ln 2 + 𝑣 2 = ln 𝑐
1 + 𝑣 2 = 𝐴 2 + 𝑣 2 + 𝐵𝑣 + 2𝑣 2 𝐶
1
2𝐴 = 1
𝐴=
2
𝐵=0
1
𝐶=
𝐴 + 2𝐶 = 1
4
4
2
2
𝑥 𝑣 2+𝑣 =𝑐
𝑥4
𝑦
𝑥
2
𝑦
2+
𝑥
2
=𝑐
𝑦 2 2𝑥 2 + 𝑦 2 = 𝑐
Example 3 ( ref i pp. 73)
Solve
𝑥 2 + 𝑦 2 𝑑𝑥 + 𝑥 2 − 𝑥𝑦 𝑑𝑦 = 0
Solution:
𝑦 = 𝑣𝑥
𝑥 2 + 𝑣𝑥
2
2
𝑑𝑦 = 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣
𝑑𝑥 + 𝑥 2 − 𝑥 𝑣𝑥
2
𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
2
𝑥 1 + 𝑣 𝑑𝑥 + 𝑥 1 − 𝑣 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
1 + 𝑣 2 𝑑𝑥 + 1 − 𝑣 𝑣 𝑑𝑥 + 𝑥 𝑑𝑣 = 0
1 + 𝑣 𝑑𝑥 + 1 − 𝑣 𝑥 𝑑𝑣 = 0
𝑑𝑥
1−𝑣
+
𝑑𝑣 = 0
𝑥
1+𝑣
𝑑𝑥
2
+
− 1 𝑑𝑣 = 0
𝑥
1+𝑣
ln 𝑥 + 2 ln 1 + 𝑣 − 𝑣 = ln 𝑐
𝑥 1 + 𝑣 2 𝑒 −𝑣 = 𝑐
𝑦
𝑥 1+
𝑥
𝑥+𝑦
2
2
𝑦
𝑒−
= 𝑐𝑥𝑒
𝑥
𝑦
=𝑐
𝑥
Example 3 ( ref i pp. 73)
Solve
𝑥 2 + 𝑦 2 𝑑𝑥 + 𝑥 2 − 𝑥𝑦 𝑑𝑦 = 0
Solution:
𝑥 = 𝑢𝑦
𝑑𝑥 = 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢
𝑢𝑦 2 + 𝑦 2 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 + 𝑢𝑦 2 − 𝑢𝑦 𝑦 𝑑𝑦 = 0
𝑦 2 𝑢2 + 1 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 + 𝑦 2 𝑢2 − 𝑢 𝑑𝑦 = 0
𝑢2 + 1 𝑢 𝑑𝑦 + 𝑦 𝑑𝑢 + 𝑢2 − 𝑢 𝑑𝑦 = 0
𝑢3 + 𝑢2 𝑑𝑦 + 𝑢2 + 1 𝑦 𝑑𝑢 = 0
𝑑𝑦
𝑢2 + 1
+ 2
𝑑𝑢 = 0
𝑦
𝑢 𝑢+1
𝑑𝑦
1 1
2
+ − + 2+
𝑑𝑢 = 0
𝑦
𝑢 𝑢
𝑢+1
ln 𝑦 − ln 𝑢 − 𝑢−1 + 2 ln 𝑢 + 1 = ln 𝑐
𝑦 𝑢+1
ln
𝑢
2
= ln 𝑐 + 𝑢 −1
𝑢2 + 1
𝐴 𝐵
𝐶
= + 2+
𝑢2 𝑢 + 1
𝑢 𝑢
𝑢+1
𝑢2 + 1 = 𝐴 𝑢2 + 𝑢 + 𝐵 𝑢 + 1 + 𝐶𝑢2
𝐵=1
𝐴+𝐵 =0
𝐴 = −1
𝐴+𝐶 =1
𝐶=2
𝑦 𝑢+1
ln
𝑢
𝑦 𝑢+1
𝑢
𝑥
𝑦 𝑦+1
𝑥
𝑦
𝑥+𝑦
2
2
= ln 𝑐 + 𝑢 −1
2
= 𝑐𝑒 𝑢
−1
2
= 𝑐𝑒
= 𝑐𝑥𝑒
𝑦
𝑦
𝑥
𝑥
𝑥 = 𝑢𝑦
𝑥
𝑢=
𝑦
Assignment #3
Exercises 1-21, 23-35 on pp. 28-29 of ref ii
Example 1 ( ref ii pp. 32)
Solution:
𝑀 = 3𝑥 2 𝑦 − 6𝑥
𝑁 = 𝑥 3 + 2𝑦
𝜕𝑀
= 3𝑥 2
𝜕𝑦
𝜕𝑁
= 3𝑥 2
𝜕𝑥
𝜕𝐹
= 3𝑥 2 𝑦 − 6𝑥
𝜕𝑥
𝐹=
𝑥 3𝑦
−
3𝑥 2
+𝑔 𝑦
𝑔′ 𝑦 = 2𝑦
𝜕𝑀 𝜕𝑁
=
𝜕𝑦
𝜕𝑥
𝜕𝐹
= 𝑥 3 + 𝑔′ 𝑦 = 𝑁
𝜕𝑦
𝑥3
+ 𝑔′ 𝑦 =
𝑥3
+ 2𝑦
𝑔 𝑦 = 𝑦2
𝐹 = 𝑥 3 𝑦 − 3𝑥 2 + 𝑦 2
𝑥 3 𝑦 − 3𝑥 2 + 𝑦 2 = 𝑐
Example 2 ( ref ii pp. 33)
Solution:
𝑀 = 2𝑥 3 − 𝑥𝑦 2 − 2𝑦 + 3
𝑁 = −𝑥 2 𝑦 − 2𝑥
𝜕𝑀
= −2𝑥𝑦 − 2
𝜕𝑦
𝜕𝑁
= −2𝑥𝑦 − 2
𝜕𝑥
𝜕𝑀 𝜕𝑁
=
𝜕𝑦
𝜕𝑥
𝜕𝐹
= −𝑥 2 𝑦 − 2𝑥
𝜕𝑦
1
𝐹 = − 𝑥 2 𝑦 2 − 2𝑥𝑦 + ℎ 𝑥
2
𝜕𝐹
= −𝑥𝑦 2 − 2𝑦 + ℎ′ 𝑥
𝜕𝑥
−𝑥𝑦 2 − 2𝑦 + ℎ′ 𝑥 = 2𝑥 3 − 𝑥𝑦 2 − 2𝑦 + 3
ℎ′ 𝑥 = 2𝑥 3 + 3
1 4
ℎ 𝑥 = 𝑥 + 3𝑥
2
1
1
𝐹 = 𝑥 4 + 3𝑥 − 𝑥 2 𝑦 2 − 2𝑥𝑦
2
2
1 4
1
𝑐
𝑥 + 3𝑥 − 𝑥 2 𝑦 2 − 2𝑥𝑦 =
2
2
2
𝑥 4 + 6𝑥 − 𝑥 2 𝑦 2 − 4𝑥𝑦 = 𝑐
Example 3 ( ref iii pp. 22)
Solution:
𝑀 = cos 𝑥 + 𝑦
𝑁 = 3𝑦 2 + 2𝑦 + cos 𝑥 + 𝑦
𝜕𝑀
= − sin 𝑥 + 𝑦
𝜕𝑦
𝜕𝑁
= − sin 𝑥 + 𝑦
𝜕𝑥
𝜕𝑀 𝜕𝑁
=
𝜕𝑦
𝜕𝑥
𝜕𝐹
= cos 𝑥 + 𝑦
𝜕𝑥
𝐹 = sin 𝑥 + 𝑦 + 𝑔 𝑦
𝐹 = sin 𝑥 + 𝑦 + 𝑦 3 + 𝑦 2
𝜕𝐹
= cos 𝑥 + 𝑦 + 𝑔′ 𝑦
𝜕𝑦
sin 𝑥 + 𝑦 + 𝑦 3 + 𝑦 2 = 𝑐
𝑔′ 𝑦 = 3𝑦 2 + 2𝑦
𝑔 𝑦 = 𝑦3 + 𝑦2
Example 4 ( ref iii pp. 23)
Solution:
𝑀 = cos 𝑦 sinh 𝑥 + 1
𝑁 = − sin 𝑦 cosh 𝑥
𝜕𝑀
= − sin 𝑦 sinh 𝑥
𝜕𝑦
𝜕𝑀 𝜕𝑁
=
𝜕𝑦
𝜕𝑥
𝜕𝑁
= − sin 𝑦 sinh 𝑥
𝜕𝑥
𝜕𝐹
= − sin 𝑦 cosh 𝑥
𝜕𝑦
𝐹 = cos 𝑦 cosh 𝑥 + ℎ 𝑥
𝜕𝐹
= cos 𝑦 sinh 𝑥 + ℎ′ 𝑥
𝜕𝑥
ℎ′ 𝑥 = 1
ℎ 𝑥 =𝑥
𝐹 = cos 𝑦 cosh 𝑥 + 𝑥
cos 𝑦 cosh 𝑥 + 𝑥 = 𝑐
@ 𝑥 = 1, 𝑦 = 2
𝑐 =0.358
cos 𝑦 cosh 𝑥 + 𝑥 = 0.358
Example 5 ( ref iv pp. 94)
Solution:
𝑀 = 2𝑥 + 𝑦 2
𝑁 = 2𝑥𝑦
𝜕𝑀
= 2𝑦
𝜕𝑦
𝜕𝑁
= 2𝑦
𝜕𝑥
𝜕𝑀 𝜕𝑁
=
𝜕𝑦
𝜕𝑥
𝜕𝐹
= 2𝑥𝑦
𝜕𝑦
𝐹 = 𝑥𝑦 2 + 𝑥 2
𝐹 = 𝑥𝑦 2 + ℎ 𝑥
𝑥𝑦 2 + 𝑥 2 = 𝑐
𝜕𝐹
= 𝑦 2 + ℎ′ 𝑥
𝜕𝑥
ℎ′ 𝑥 = 2𝑥
ℎ 𝑥 = 𝑥2
Example 6 ( ref iv pp. 97)
Solution:
𝑀 = 𝑦 cos 𝑥 + 2𝑥𝑒 𝑦
𝑁 = sin 𝑥 + 𝑥 2 𝑒 𝑦 − 1
𝜕𝑀
= cos 𝑥 + 2𝑥𝑒 𝑦
𝜕𝑦
𝜕𝑀 𝜕𝑁
=
𝜕𝑦
𝜕𝑥
𝜕𝐹
= 𝑦 cos 𝑥 + 2𝑥𝑒 𝑦
𝜕𝑥
𝜕𝑁
= cos 𝑥 + 2𝑥𝑒 𝑦
𝜕𝑥
𝐹 = 𝑦 sin 𝑥 + 𝑥 2 𝑒 𝑦 + 𝑔 𝑦
𝜕𝐹
= sin 𝑥 + 𝑥 2 𝑒 𝑦 + 𝑔′ 𝑦
𝜕𝑦
𝑔′ 𝑦 = −1
𝑔 𝑦 = −𝑦
𝐹 = 𝑦 sin 𝑥 + 𝑥 2 𝑒 𝑦 − 𝑦
𝑦 sin 𝑥 + 𝑥 2 𝑒 𝑦 − 𝑦 = 𝑐
Example 1 ( ref ii pp. 32)
Solution:
𝑀 = 3𝑥 2 𝑦 − 6𝑥
𝑁 = 𝑥 3 + 2𝑦
𝜕𝐹
= 3𝑥 2 𝑦 − 6𝑥
𝜕𝑥
𝐹 = 𝑥 3 𝑦 − 3𝑥 2 + 𝑔 𝑦
𝜕𝐹
= 𝑥 3 + 2𝑦
𝜕𝑦
𝐹 = 𝑥 3𝑦 + 𝑦 2 + ℎ 𝑥
𝑔 𝑦 = 𝑦2
ℎ 𝑥 = −3𝑥 2
𝐹 = 𝑥 3 𝑦 − 3𝑥 2 + 𝑦 2
𝑥 3 𝑦 − 3𝑥 2 + 𝑦 2 = 𝑐
Example 2 ( ref ii pp. 33)
Solution:
𝑀 = 2𝑥 3 − 𝑥𝑦 2 − 2𝑦 + 3
𝑁 = −𝑥 2 𝑦 − 2𝑥
𝜕𝐹
= 2𝑥 3 − 𝑥𝑦 2 − 2𝑦 + 3
𝜕𝑥
1
1
𝐹 = 𝑥 4 − 𝑥 2 𝑦 2 + 2𝑥𝑦 + 3𝑥 + 𝑔 𝑦
2
2
𝜕𝐹
= −𝑥 2 𝑦 − 2𝑥
𝜕𝑦
1 2 2
𝐹 = − 𝑥 𝑦 − 2𝑥𝑦 + ℎ 𝑥
2
1
ℎ 𝑥 = 𝑥 4 + 3𝑥
2
𝑔 𝑦 =0
1 4 1 2 2
𝑥 − 𝑥 𝑦 + 2𝑥𝑦 + 3𝑥 + 𝑔 𝑦
2
2
1 4 1 2 2
𝑐
𝑥 − 𝑥 𝑦 + 2𝑥𝑦 + 3𝑥 =
2
2
2
𝐹=
𝑥 4 + 6𝑥 − 𝑥 2 𝑦 2 − 4𝑥𝑦 = 𝑐