: Integrating Factors to Orthogonal Trajectories of Curves

1. ACTIVE LEARNING ASSIGNMENT
SUBJECT : Applied Engineering Mathematics
TOPIC : Integrating Factors to Orthogonal Trajectories of Curves
Group No. : 6
Branch : Electrical Engineering

2. Content
 Integrating factors
 Linear differential equations
 Equations reducible to linear from Bernoulli Equation
 Orthogonal Trajectories
 Working Rule for Finding Orthogonal Trajectories

3. Integrating factors
 Sometimes we have an equation
P (x, y) dx + Q (x, y) dy = 0 …..(1)
that is not exact, but if we multiply it by a suitable function F(x, y), the new
equation
F P dx + F Q dy = 0 …..(2)
becomes exact, so it can be solved by the method of Exact differential
equation method. The function F(x, y) is then called an Integrating factor of
(1).

4. • Examples :-
1. Verify that F = 1/y2 is an integrating factors of y (2xy + ex) dx – ex dy = 0
then find the general solution.
Solution :- Multiplying by F = 1/y2 gives the new equation
2x dx + (y ex dx – ex dy)/(y2) = 0
d (x2) + d (ex/y) = 0
and its solution is x2 + (ex/y) = c i.e. x2y + ex = cy

5. a) Integrating factors found by Inspection
 In simpler cases, integrating factors may be found by inspection or perhaps
after some trials. The following differentials are useful in selecting a suitable
integrating factors.
Differentials
I. x dy + y dy + d(xy)
II. (x dy – y dy)/(x2) = d(y/x)
III. (y dx – x dy) /(y2) = d (x/y)
IV. (x dy + y dx)/(xy) = d [ log (xy)]
V. (x dy – y dx)/(xy) = d [ log (y/x)]
VI. (y dx – x dy)/(xy) = d [ log (x/y)]
VII. (x dy – y dx)/(x2 + y2) = d [tan-1(y/x)]
VIII. (x dx + y dy)/(x2 + y2) = ½ d [ log (x2 + y2)]
IX. (x dy – y dx )/(x2 – y2) = d(1/2 log (x + y)/(x – y))

6. Examples :-
1. Solve y dx – x dy + (1 + x2) dx + x2 sin y dy = 0
Solution . Dividing by x2
(y dx – x dy)/(x2) + (1/x2 + 1) dx + sin y dy = 0
d(-y/x) + d (-1/x + x) + d (-cos y) = 0 which is exact.
Integrating we get
-y/x -1/x + x – cos y = c
i.e. y + 1 –x2 + x cos y + cx = 0
which is the required solution.

7. b)Rule for finding Integrating factors
Rule 1 : If Mx + Ny ≠ 0 and the equation is homogeneous, then 1/(Mx + Ny)
Is an integrating factor.
Example: Solve (xy – 2y2) dx – (x2 – 3xy) dy = 0
Solution: The given equation is homogeneous in x and y.
with M = xy – 2y2, N = -x2 + 3xy, (∂M/∂y ≠ ∂N/∂y)
Mx + Ny = x2y – 2xy2 – x2y + 3xy2 = xy2 ≠ 0
I.F. = 1/(Mx + Ny) = 1/xy2
Multiplying throughout by 1/xy2, the given equation becomes
(1/y – 2/x)dx + (-x/y2 + 3/y)dy = 0, which is exact, hence solving as usual
the solution is x/y – 2 log x + 3 log y = c

8. RULE 2 :- If Mx – Ny ≠ 0 and the equation can be written in the form f1(xy) y dx
+ f2(xy) x dy = 0 then 1/(Mx – Ny) is an integrating factor.
Example:- Solve the differential equation :
(x2y2 + 2) y dx + (2 – x2y2)mx dy = 0
Solution:- The equation is of the form
f1(xy) y dx + f2(xy) x dy = 0
Here M = x2y3 + 2y, N = 2x – x3y2
Now Mx – Ny = x3y3 + 2xy – 2xy + x3y3 = 2x3y3 ≠ 0
I.F. = 1/(Mx –Ny) = 1/2x3y3
Multiplying the equation by 1/2x3y3 we have
(x2y2 + 2) y ∙ 1/(2x3y3) dx + (2 – x2y2) x ∙ 1/(2x3y3) dy = 0
Or ½ (1/x + 2/x3y2) dx + ½ (2/x2y3 - 1/y) dy = 0 which is exact
∫ M dx = ½ ∫ (x-1 + 2y-2 x-3) dx = ½ (log x – 1/x2y2)
∫ N dy = -1/2 ∫1/y dy = -1/2 log y
The solution is
½ ( log x – 1/ x2y2) – ½ log y = constant
Or log (x/y) – 1/(x2y2) = C

9. RULE 3 :- For the equation Mdx + Ndy = 0
If 1/N(∂M/∂y - ∂N/∂x) is a function of x alone , say f(x), then e∫f(x) dx is an
integrating factor.
Example:- Solve (x2 + y2 + 1) dx – 2xy dy = 0
Solution:- Here M = x2 + y2 + 1 and N = -2xy
1/N(∂M/∂y - ∂N/∂x) = -1/2xy (2y + 2y)
= -2/x which is a function of x only
I.F. = e∫-2/x dx = e-2 log x
= elog x-2
= x-2
Multiplying the equation by 1/x2, we have
(1 + y2/x2 + 1/x2) dx – 2y/x dy = 0 which is a exact
∫ Mdx = ∫ (1 + y2/x2 + 1/x2) dx and ∫ Ndy = 0
= x- y2/x -1/x2
The solution is x- y2/x -1/x2= C

10. RULE 4:- For the equation Mdx + Ndy = 0
If 1/M(∂N/∂x - ∂M/∂y) is a function of y alone , say g(y), then e∫g(y) dx is an
integrating factor.
Example:- Solve (xy3 + y) dx + 2 (x2y2 + x + y4) dy = 0
Solution:- Here M = xy3 + y and N = 2 x2y2 + 2x + 2y4
∂M/∂y = 3xy2 + 1, ∂N/∂x = 4xy2 + 2
Now 1/M(∂N/∂x - ∂M/∂y) = {(4xy2 + 2) – (3xy2 + 1)}/y(xy2 + 1)
= (xy2 + 1)/y(xy2 + 1)
= 1/y, which is a function of y only.
I.F. = e∫1/y dy = elog y = y
Multiplying the equation by y, we have
(xy4 + y2) dx + 2 (x2y3 + xy +y5)dy = 0 which is exact
∫ Mdx = ∫ (xy4 + y2) dx = (x2y4/2) + xy2 and
∫ Ndy = ∫ 2y5 dy = y6/3
The solution is (x2y4)/2 + xy2 + y6/3 = C

11. Linear Differential Equation
When a differential equation is said to be linear if the dependent variable and
its derivative occur only in the first degree and are not multiplied together.
The form of the linear equation of the first order is dy/dx + Py = Q…...(1)
Where P and Q are function of x or constant.
Equation (1) is also known as Leibnitz’s linear equation.
Method of Solution :
Writing equation (1) as
(Py – Q) dx + dy = 0
Where M = Py – Q and N = 1
∂M/∂y = P, ∂N/∂x = 0
Hence 1/N(∂M/∂y - ∂N/∂x) = 1/1 (P – 0) = P(x)
I.F. = e∫Pdx
Multiplying the equation (1) by e∫Pdx , we get
e∫Pdx (dy/dx + Py) = Q e∫Pdx

12. d/dx (y e∫Pdx) = Q e∫Pdx
Integrating both the side with respect to x, we have
Y e∫Pdx = ∫ Q e∫Pdx dx + c….(2)
Which is the required general solution of the given linear equation, which can
be written as
y (I.F.) dx + c
Similarly dx/dy + Px = Q, where P and Q are function of y or constant, is a
linear differential equation with x as dependent and y as the independent
variable.
Integrating factor in this case is e∫Pdy and the solution is
x (I.F) = ∫ Q (I.F.) dy + c.

13. Example :- Solve x dy/dx – ay = x + 1
Solution:- The given equation can be written as
dy/dx – ay/x = (x + 1)/x
Comparing it with dy/dx + Py = Q, we have
P = -a/x, Q = (x + 1)/x
I.F. = e∫ Pdx = e∫ -a/x dx = e-a log x
= elog x-a =1/xa
The solution is
y (I.F.) = ∫ Q (I.F.) dx + c
y 1/xa = ∫ (x + 1)/x ∙ 1/xa dx + c
= ∫ (1/xa + 1/xa+) dx + c
y ∙ 1/xa = x1-a/1-a – x-a/a +c
y = x/(1-a) – 1/a + cxa is required solution

14. Bernoulli Equation
Certain nonlinear differential equation can be reduced to linear form. The
most famous of these is the Bernoulli Equation
a) An equation of the form
dy/dx + Py = Q yn ……….(1)
Where P and Q are function of x or constant and n any number is called
Bernoulli’s equation.
If n = 0 or n = 1, the equation is linear. Otherwise it is non linear.
Dividing both side of (1) by yn, we get
y-n dy/dx + Py1-n = Q………..(2)
Putting y1-n = v, (1-n)y-n dy/dx = dv/dx
Equation (2) becomes
1/(1-n) dv/dx + Pv = Q
Or dv/dx + (1-n) Pv = (1-n) Q
Which is a linear differential equation in v and x

15. b) Equation of the form reducible to linear form is
f’(y) dy/dx + P f(y) = Q
Where P and Q are function of x or constant
Putting f(y) = v so that f’(y) dy/dx = dv/dx
Equation reduced to
dv/dx + Pv = Q, which is linear
Example:- Solve dy/dx = 2y tan x + y2 tan2x
Solution:- The given equation can be written as
dy/dx - 2y tan x = y2 tan2x
Dividing by y2, we have
y-2 dy/dx – 2y-1 tan x = tan2x
Putting –y-1 = v so that y-2 dy/dx =dv/dx
Therefore equation reduces to
dv/dx + (2 tan x) v = tan2x which is linear in v and x
I.F. = e∫Pdx = e∫2 tan x dx = e2 log sec x = sec2x
The solution is………..

16. v (I.F.) = ∫ (tan2x) (I.F) dx + c
v (sec2x) = ∫ (tan2x) sec2x dx + c
= 1/3 tan3x + c
-1/y sec2x = 1/3 tan3x +c (v = -1/y)
Or 1/y sec2x = c – 1/3 tan3x

17. Orthogonal Trajectories
In many engineering applications, we use differential equations for finding
curves that intersect given curves at right angles. The new curves are then called
the orthogonal trajectories of the curves and vice versa. Here “Orthogonal” is
another word for “Perpendicular”. Refer fig.
For instance, in two- dimensional
problems in the flow of heat, the lines of heat
flow in a body are everywhere perpendicular to
the isothermal curves. The meridians and
parallels on the earth are orthogonal trajectories
of each other. In an electric field, the curves of
electric force are the orthogonal trajectories of
the equipotential lines (curves of constant
voltage) and conversely. In fluid flow, stream
lines and equipotential lines cut orthogonally.
Other important examples arise In
hydrodynamics, heat conduction and other fields
of engineering.

18. • One- parameter family of curves : If for each fixed value of c the equation f
(x, y, c) = 0 represents a curve in the xy-plane and if for various value of c it
represents infinite number of curves, then the totality of these curves is called
a one-parameter family of curves, and c is called the parameter of the
family.
Working Rule for Finding Orthogonal Trajectories
(a) For Cartesian curves f(x, y, c) = 0
Step 1 Given a family of curves f(x, y, a) = 0 . . . . (1)
where a is a parameter.
Step 2 Differentiate (1) w.r.t.x. and eliminate “a” between (1)
and the resulting equation. We thus form a differential
equation of the given family of the form
F (x, y, dy/dx) = 0 . . . . (2)
Step 3 Replace dy/dx by – 1/(dy/dx) in (2). Then the differential
equation of the orthogonal trajectories will be
F (x, y, -dy/dx) = 0 . . . . (3)

19. Step 4 Solve equation (3) to get equation of the required orthogonal
trajectories of the form
F (x, y, b) = 0 . . . . (4)
For polar curves f(r, θ, c) = 0
Step 1 Given a family of curves f(r, θ, a) = 0 . . . . (1)
where a is a parameter.
Step 2 Differentiate (1) w.r.t. θ and eliminate ‘a’ between (1) and the
resulting equation. We thus form a differential equation of the
given family of the form.
F (r, θ, dr/dθ) = 0 . . . . (2)
Step 3 Replace dr/dθ by –r2(dθ/dr) or 1/r (dr/dθ) by –r (dθ/dr) in (2). Then the
differential equation of the orthogonal trajectories will be
F( r, θ, -r2 dθ/dr) = 0 . . . . (3)
Step 4 Solve equation (4) to get equation of the required orthogonal
trajectories of the form
F (r, θ, b) = 0 . . . . (4)

20. EXAMPLE:- Find the orthogonal trajectories of the curve y = x2 + c.
SOLUTION
y = x2 + c Eqn. of the given trajectory
dy/dx = 2x Diffl. Eqn. of the given trajectory
dy/dx = -1/2x Diffl. Eqn. of O.T.
dy = -(1/2x)dx solving, we get
y = -1/2 logx + loga = log(a/√x)
Or a/√x = ey or √x ey = a
Is required orthogonal trajectory.

21. EXAMPLE : Find the orthogonal trajectories of the cardioids.
r = a (1 + cosθ)
SOLUTION r = a (1+ cos θ) . . . . (1)
Where a is a parameter is the equation of the given family of cardioid.
Taking logarithm on both sides.
log r = log a + log (1 + cosθ)
Differentiating w.r.t. θ. We get
(1/r) dr/dθ = 0 + 1/(1+cosθ)x (-sinθ)
= - [2sin(θ/2)cos(θ/2)] / 2cos2 (θ/2)
i.e. (1/r)(dr/dθ) = tan θ/2 . . . . (2)
Which is the differential equation of the given family. (1).

22. Replacing (1/r)(dr/dθ) by -r (dθ/dr) in (2) , we get
(1/r) dr = cot(θ/2) dθ . . . .(3)
Which is the differential equation of the orthogonal trajectories.
Integrating we get
log r = 2 log sin (θ/2) + log c
log r = log c sin2(θ/2)
R = c sin2(θ/2) = c/2 (1 - cosθ)
or
So,
Writing b for c/2, we get
r = b (1 - cos θ) (fig. 2) . . . .(4)
The required orthogonal trajectories.

23. Thank You