ACTIVE LEARNING ASSIGNMENT

SUBJECT : Applied Engineering MathematicsTOPIC : Integrating Factors to Orthogonal Trajectories of Curves Group No. : 6Branch : Electrical Engineering

Content

Integrating factors

Linear differential equations

Equations reducible to linear from Bernoulli Equation

Orthogonal Trajectories

Working Rule for Finding Orthogonal Trajectories

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).

• 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

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.

DifferentialsI. 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))

Examples :-

1. Solve y dx – x dy + (1 + x2) dx + x2 sin y dy = 0Solution . 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.

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 = 0Solution: 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

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 = 0Here M = x2y3 + 2y, N = 2x – x3y2Now 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 = 0Or 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

RULE 3 :- For the equation Mdx + Ndy = 0If 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 = 0Solution:- 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

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 = 0Solution:- Here M = xy3 + y and N = 2 x2y2 + 2x + 2y4

∂M/∂y = 3xy2 + 1, ∂N/∂x = 4xy2 + 2Now 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 = yMultiplying 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 = y6/3The solution is (x2y4)/2 + xy2 + y6/3 = C

Linear Differential EquationWhen 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

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.

Example :- Solve x dy/dx – ay = x + 1 Solution:- The given equation can be written as dy/dx – ay/x = (x + 1)/xComparing it with dy/dx + Py = Q, we have P = -a/x, Q = (x + 1)/xI.F. = e∫ Pdx = e∫ -a/x dx = e-a log x

= elog x-a =1/xaThe solution is y (I.F.) = ∫ Q (I.F.) dx + cy 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 +cy = x/(1-a) – 1/a + cxa is required solution

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/dxEquation (2) becomes 1/(1-n) dv/dx + Pv = Q Or dv/dx + (1-n) Pv = (1-n) QWhich is a linear differential equation in v and x

b) Equation of the form reducible to linear form is f’(y) dy/dx + P f(y) = QWhere P and Q are function of x or constant Putting f(y) = v so that f’(y) dy/dx = dv/dxEquation reduced to dv/dx + Pv = Q, which is linear

Example:- Solve dy/dx = 2y tan x + y2 tan2xSolution:- 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 = tan2xPutting –y-1 = v so that y-2 dy/dx =dv/dxTherefore 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 = sec2xThe solution is………..

v (I.F.) = ∫ (tan2x) (I.F) dx + cv (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

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.

• 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 differentialequation of the given family of the formF (x, y, dy/dx) = 0 . . . . (2)

Step 3 Replace dy/dx by – 1/(dy/dx) in (2). Then the differentialequation of the orthogonal trajectories will be F (x, y, -dy/dx) = 0 . . . . (3)

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) = 0Step 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 theresulting 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 orthogonaltrajectories of the formF (r, θ, b) = 0 . . . . (4)

EXAMPLE:- Find the orthogonal trajectories of the curve y = x2 + c.

SOLUTIONy = 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.

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).

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 clog r = log c sin2(θ/2)R = c sin2(θ/2) = c/2 (1 - cosθ)

orSo,

Writing b for c/2, we get r = b (1 - cos θ) (fig. 2) . . . .(4)

The required orthogonal trajectories.

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