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...


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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A course in ordinary differential equations by Stephen A. Wirkus, Visit Amazon's Randall J. Swift Page, search results, Learn about Author Central, Randall J. Swift,

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