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AEM Integrating factor to orthogonal trajactories

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: Integrating Factors to Orthogonal Trajectories of Curves

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AEM Integrating factor to orthogonal trajactories

  1. 1. ACTIVE LEARNING ASSIGNMENT SUBJECT : Applied Engineering Mathematics TOPIC : Integrating Factors to Orthogonal Trajectories of Curves Group No. : 6 Branch : Electrical Engineering
  2. 2. Content  Integrating factors  Linear differential equations  Equations reducible to linear from Bernoulli Equation  Orthogonal Trajectories  Working Rule for Finding Orthogonal Trajectories
  3. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 23. Thank You

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