VTU Engg Maths-I

1. BHEEMANNA KHANDRE INSTITUTE OF TECHNOLOGY, BHALKI
Dept. of Applied Science and Humanities, (Mathematics)
Sub: Engg. Mathematics-I
Sub.code: 15MAT11
Power Point Presentation of Module -IV
by
Dr. Jagadish Tawade

2. Syllabus of Module -IV
Differential equations and its applications:
Solutions of ordinary differential equations of first order and first degree:-Bernoulli’s
equations, Exact differential equations, equations reducible to exact equations:-
Integrating factor of homogeneous equation, Integrating factor of the equation
f1(x,y).y dx + f2(x,y).x dy = 0, Integrating factor of the equation Mdx + Ndy = 0
Applications of Differential equations:
Orthogonal trajectories, Newton’s law of cooling. Flow of electricity and laws of
decay and growth.
[8hrs]

3. Ordinary Differential Equations (1 of 2)
 A differential equation is an algebraic
equation that contains some derivatives:
03735 2
2
=++=+ y
dx
dy
dx
yd
ty
dt
dy
• Recall that a derivative indicates a change in a
dependent variable with respect to an
independent variable.
• In these two examples, y is the dependent
variable and t and x are the independent
variables, respectively.

4. Why study differential equations?
• Many descriptions of natural phenomena are
relationships (equations) involving the rates at
which things happen (derivatives).
• Equations containing derivatives are called
differential equations.
• Ergo, to investigate problems in many fields of
science and technology, we need to know
something about differential equations.

5. Why study differential equations?
• Some examples of fields using differential
equations in their analysis include:
— Solid mechanics & motion
— heat transfer & energy balances
— vibrational dynamics & seismology
— aerodynamics & fluid dynamics
— electronics & circuit design
— population dynamics & biological systems
— climatology and environmental analysis
— options trading & economics

6. Examples of Fields Using Differential
Equations in Their Analysis
Cool
Bath
Hot
Bath
ky
dt
dy
ckx
dt
dx
c
dt
xd
m +=++2
2
02
2
=++ kx
dt
dx
c
dt
xd
m
ikkx
dt
dx
c
dt
xd
m f+−−=2
2
ghAC
dt
dh
A od 2−=
( )70−−= Hk
dt
dH

7. Differential Equation Basics
• The order of the highest derivative in
a differential equation indicates the order of
the equation.
D.E.PartialOrderSecond0
1
EquationOrderSecond037
EquationOrderFirst35
2
2
2
2
=
∂
∂
−
∂
∂
=++
=+
t
T
x
T
y
dx
dy
dx
yd
ty
dt
dy
α

8. Simple Differential Equations
A simple differential equation has the
form
( )
dy
f x
dx
=
Its general solution is
( )y f x dx= ∫

9. Ex. Find the general solution to
Simple Differential Equations
( )dxxxdy 32
42 −=
32
42 xx
dx
dy
−=
( )∫ −= dxxxy 32
42
Cxxy +−= 43
3
2

10. Ex. Find the general solution to
Simple Differential Equations
dxx
x
dy 





−= 2
1
x
xdx
dy
2
1
−=
∫ 





−= dxx
x
y 2
1
Cxxy +−= 2
ln

11. Find the general solution to
Exercise: (Waner, Problem #1, Section 7.6)
( )dxxxdy += 2
xx
dx
dy
+= 2
( ) ( )∫∫ +=+= dxxxdxxxy 2122 /
Cxxy ++= 233
3
2
3
1 /

12. A drag racer accelerates from a stop so that
its speed is 40t feet per second t seconds after
starting. How far will the car go in 8
seconds?
Example: Motion
seconds.intimeisand
feet,indistancetheis)(wher,40
t
tst
dt
ds
=
( ) ft?8 =s
Given:
Find:

13. dttds 40=
t
dt
ds
40=
( ) ( ) Ctdttts +== ∫
2
2040
( ) ( ) ft12808208
2
==s
Solution:
Apply the initial condition: s(0) = 0
( ) ( ) Cs +==
2
02000 0=C
( ) 2
20tts =
The car travels 1280 feet in 8 seconds

14. Find the particular solution to
( )dxxxdy 23
−=
0when1;23
==−= xyxx
dx
dy
( )∫ +−=−= Cxxdxxxy 243
4
1
2
1
4
1 24
+−= xxy
Exercise: (Waner, Problem #11, Section 7.6)
Apply the initial condition: y(0) = 1
( ) ( ) C+−=
24
00
4
1
1 1=C

15. Separable Differential Equations
A separable differential equation has the form ( ) ( )
dy
f x g y
dx
=
Its general solution is
1
( )
( )
dy f x dx
g y
=∫ ∫
Consider the differential equation 2
y
x
dx
dy
=
Example: Separable Differential Equation
a. Find the general solution.
b. Find the particular solution that satisfies
the initial condition y(0) = 2.

16. dxxdyy =2
Solution:
Step 1 — Separate the variables:
∫∫ = dxxdyy2
Step 2 — Integrate both sides:
Step 3 — Solve for the dependent variable:
C
xy
+=
23
23
a.
31
2
31
2
2
3
3
2
3
//






+=





+= DxCxy
This is the general solution

17. Solution: (continued)
Apply the initial (or boundary) condition, that is,
substituting 0 for x and 2 for y into the general
solution in this case, we get
( ) 31
31
2
0
2
3
2 /
/
DD =





+=
Thus, the particular solution we are looking for is
823
==D
b.
31
2
8
2
3
/






+= xy

18. Find the general solution to
Exercise: (Waner, Problem #4, Section 7.6)
x
dx
y
dy
=
x
y
dx
dy
=
∫∫ =
x
dx
y
dy
)(where C
eAAxy ±==
Cxy += lnln
xeey CCx
== +ln
Axxey C
=±=