By Stephen A. Wirkus, Visit Amazon's Randall J. Swift Page, search results, Learn about Author Central, Randall J. Swift,

ISBN-10: 1420010417

ISBN-13: 9781420010411

ISBN-10: 1584884762

ISBN-13: 9781584884767

"Featuring real-world purposes from engineering and technology fields, A path in traditional Differential Equations is the 1st ebook on traditional differential equations (ODEs) to incorporate suitable desktop code and directions of MATLAB®, Mathematica®, and Maple. The ebook embeds the pc algebra code all through, offering the syntax subsequent to the correct concept. It absolutely describes approximations used to obtain Read more...

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that includes real-world functions from engineering and technological know-how fields, this booklet on usual differential equations (ODEs) comprises correct machine code and directions of MATLAB[registered], Read more...

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Example text

27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. y 2 +2xy x2 dy 2x2 dx = x2 + dy dx = y2 xy − y = x2 + y 2 (x + 2y)dx − xdy = 0 (y 2 − 2xy)dx + x2 dy = 0 2x3 y = y(2x2 − y 2 ) (x2 + y 2 )y = 2xy xy − y = x tan( xy ) (2x + y)dx − (4x + 2y)dy = 0 y 2 + x2 y = xyy x − y + (y − x)y = 0 (x + 4y)y = 2x + 3y (x − y)dx + (x + y)dy = 0 ydx = (2x + y)dy y y = 2( x+y )2 39. 2xdy + (x2 y 4 + 1)ydx = 0 40. ydx + x(2xy + 1)dy = 0 41. A function F is called homogeneous of degree n if F (tx, ty) = tn F (x, y) for all x and y.

The intersection of y − x + 1 = 0 and y + x + 5 = 0 is (−2, −3). Check it! Let x = X − 2 and y = Y − 3. This amounts to taking new axes parallel to the old with an origin at (−2, −3). (ii) Use this transformation to obtain the differential equation Y −X dY = . 5). Use the technique of problem 45 to solve problems 46–50. 46. (2x + y + 1)dx − (4x + 2y − 3)dy = 0 47. x − y − 1 + (y − x + 2)y = 0 48. (x + 4y)y = 2x + 3y − 5 49. (y + 2)dx = (2x + y − 4)dy y+2 2 50. 3 Some Physical Problems Arising as Separable Equations Now that we have studied separable equations in detail, we consider some applications.

Example 2: Solve dx = et−x , x(0) = ln 2, dt for x(t). Separating the variables gives dx = et e−x dt and thus ex dx = et dt. Integrating both sides of this equation gives ex = et + C. Solving for x, we have x = ln |et + C|. Applying the initial condition x(0) = ln 2 yields ln 2 = ln |1 + C|, so that C = 1. 2. SEPARABLE DIFFERENTIAL EQUATIONS 13 Thus x = ln(et + 1), which is defined for all t. Note that et + 1 is always positive so that we can drop the absolute value signs. We should also note that after integrating, we could have applied the initial condition to determine C and then proceeded to solve for x instead of first solving for x and then applying the initial condition to determine C.

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