Chapter First Order Differential Equations Contents. The Method of Quadrature.......... 68. Separable Equations............... 74.3 Linear Equations I................ 85.4 Linear Equations II............... 93.5 Linear Applications................6 Kinetics...................... 3.7 Logistic Equation................ 3.8 Science and Engineering Applications.... 36.9 Exact Equations and Level Curves...... 5. Special equations................. 57 The subject of the chapter is the first order differential equation y = fx, y). The study includes closed-form solution formulas for special equations, numerical solutions and some applications to science and engineering.. The Method of Quadrature The method of quadrature refers to the technique of integrating both sides of an equation, hoping thereby to extract a solution formula. The name quadrature originates in geometry, where quadrature means finding area, a task overtaken in modern mathematics by integration. The naming convention is obeyed by maple, which lists it as its first method for solving differential equations. Below, Theorem, proved on page 7, isolates the requirements which make this method successful.
. The Method of Quadrature 69 Theorem Quadrature) Let F x) be continuous on a < x < b. Assume a < x < b and < y <. Then the initial value problem ) y = F x), yx ) = y has the unique solution ) x yx) = y + F t)dt. x To apply the method of quadrature means: i) Calculate a candidate solution formula by the working rule below; ii) Verify the solution. To solve y = fx, y) when f is independent of y, integrate on variable x across the equation. River Crossing A boat crosses a river at fixed speed with power applied perpendicular to the shoreline. Is it possible to estimate the boat s downstream location? The answer is yes. The problem s variables are x Distance from shore, w Width of the river, y Distance downstream, v b Boat velocity dx/dt), t Time in hours, v r River velocity dy/dt). The calculus chain rule dy/dx = dy/dt)/dx/dt) is applied, using the symbols v r and v b instead of dy/dt and dx/dt, to give the model equation 3) dy dx = v r v b. Stream Velocity. The downstream river velocity will be approximated by v r = kxw x), where k > is a constant. This equation gives velocity v r = at the two shores x = and x = w, while the maximum stream velocity at the center x = w/ is see page 7) 4) v c = kw 4. Special River-Crossing Model. The model equation 3) using v r = kxw x) and the constant k defined by 4) give the initial value problem 5) dy dx = 4v c xw x), y) =. v b w
7 First Order Differential Equations The solution of 5) by the method of quadrature is y = 4v c v b w 3 x3 + ) 6) wx, where w is the river s width, v c is the river s midstream velocity and v b is the boat s velocity. In particular, the boat s downstream drift on the opposite shore is 3 wv c/v b ). See Technical Details page 7. Examples Example Quadrature) Solve y = 3e x, y) =. Solution: Candidate solution. The working rule is applied. y t) = 3e t Copy the equation, x replaced by t. x y t)dt = x 3et dt Integrate across t x. yx) y) = 3e x 3 Fundamental theorem of calculus, page 68. yx) = 3e x 3 Candidate solution found. Used y) =. Verify solution. Let y = 3e x 3. The initial condition y) = follows from e =. To verify the differential equation, the steps are: LHS = y Left side of the differential equation. = 3e x 3) Substitute the expression for y. = 3e x Sum rule, constant rule and e u ) = u e u. = RHS Solution verified. Example River Crossing) A boat crosses a mile-wide river at 3 miles per hour with power applied perpendicular to the shoreline. The river s midstream velocity is miles per hour. Find the transit time and the downstream drift to the opposite shore. Solution: The answers, justified below, are minutes and /9 miles. Transit time. This is the time it takes to reach the opposite shore. The layman answer of minutes is correct, because the boat goes 3 miles in one hour, hence mile in /3 of an hour, perpendicular to the shoreline. Downstream drift. This is the value y), where y is the solution of equation 5), with v c =, v b = 3, w =, all distances in miles. The special model is dy dx = 4 x x), y) =. 3 3 x3 + x) and the downstream The solution given by equation 6) is y = 4 3 drift is then y) = /9 miles. This answer is /3 of the layman s answer of /3)) miles; the explanation is that the boat is pushed downstream at a variable rate from to miles per hour.
. The Method of Quadrature 7 Details and Proofs Proof of Theorem : Uniqueness. Let yx) be any solution of ). It will be shown that yx) is given by the solution formula ). yx) = y) + x x y t)dt Fundamental theorem of calculus, page 68. = y + x x F t)dt Use ). Verification of the Solution. Let yx) be given by solution formula ). It will be shown that yx) solves initial value problem ). y x) = y + x x F t)dt) Compute the derivative from ). = F x) Apply the fundamental theorem of calculus. The initial condition is verified in a similar manner: yx ) = y + x x F t)dt Apply ) with x = x. = y The integral is zero: a F x)dx =. a The proof is complete. Technical Details for 4): The maximum of a continuously differentiable function fx) on x w can be found by locating the critical points i.e., where f x) = ) and then testing also the endpoints x = and x = w. The derivative f x) = kw x) is zero at x = w/. Then fw/) = kw /4. This value is the maximum of f, because f = at the endpoints. Technical Details for 6): Let a = 4v c v b w. Then y = y) + x y t)dt Method of quadrature. = + a x tw t)dt By 5), y = atw t). = a 3 x3 + wx). Integral table. To compute the downstream drift, evaluate yw) = a w3 6 or yw) = w 3 Exercises. v c v b. Quadrature. Find a candidate solution for each initial value problem and verify the solution. See Example, page 7.. y = 4e x, y) =.. y = e 4x, y) =. 3. + x)y = x, y) =. 4. x)y = x, y) =. 5. y = sin x, y) =. 6. y = cos x, y) =. 7. y = xe x, y) =. 8. y = xe x, y) =. 9. y = tan x, y) =.. y = + tan x, y) =.
7 First Order Differential Equations. + x )y =, y) =.. + 4x )y =, y) =. 3. y = sin 3 x, y) =. 4. y = cos 3 x, y) =. 5. + x)y =, y) =. 6. + x)y =, y) =. 7. + x) + x)y =, y) =. 8. + x)3 + x)y = 3, y) =. 9. y = sin x cos x, y) =.. y = + cos x) sin x, y) =. River Crossing. A boat crosses a river of width w miles at v b miles per hour with power applied perpendicular to the shoreline. The river s midstream velocity is v c miles per hour. Find the transit time and the downstream drift to the opposite shore. See Example, page 7, and the details for 6).. w =, v b = 4, v c =. w =, v b = 5, v c = 5 3. w =., v b = 3, v c = 3 4. w =., v b = 5, v c = 9 5. w =.5, v b = 7, v c = 6 6. w =, v b = 7, v c = 7. w =.6, v b = 4.5, v c = 4.7 8. w =.6, v b = 5.5, v c = 7 Fundamental Theorem I. Verify the identity. Use the fundamental theorem of calculus part b), page 68. 9. x + t)3 dt = 4 3. x + t)4 dt = 5 + x) 4 ). + x) 5 ). 3. x te t dt = xe x e x +. 3. x tet dt = xe x e x +. Fundamental Theorem II. Differentiate. Use the fundamental theorem of calculus part b), page 68. 33. x t tant 3 )dt. 34. 3x t 3 tant )dt. 35. sin x te t+t dt. 36. sin x ln + t 3 )dt. Fundamental Theorem III. Integrate fx)dx. Use the fundamental theorem of calculus part a), page 68. Check answers with computer or calculator assist. Some require a clever u-substitution or an integral table. 37. fx) = xx ) 38. fx) = x x + ) 39. fx) = cos3πx/4) 4. fx) = sin5πx/6) 4. fx) = + x 4. fx) = x + x 4 ) 43. fx) = x e x3 44. fx) = xsinx ) + e x ) 45. fx) = 46. fx) = 47. fx) = 48. fx) = 49. fx) = 5. fx) = + x x + x + 4x x + x 4x 4x
. The Method of Quadrature 73 5. fx) = cos x sin x 5. fx) = cos x sin 3 x 53. fx) = ex + e x 54. fx) = ln x x 55. fx) = sec x 56. fx) = sec x tan x 57. fx) = csc x 58. fx) = csc x cot x 59. fx) = csc x cot xx 6. fx) = sec x tan xx Integration by Parts. Integrate fx)dx by parts, udv = uv vdu. Check answers with computer or calculator assist. 6. fx) = xe x 6. fx) = xe x 63. fx) = ln x 64. fx) = x ln x 65. fx) = x e x 66. fx) = + x)e x 67. fx) = x cosh x 68. fx) = x sinh x 69. fx) = x arctanx) 7. fx) = x arcsinx) Partial Fractions. Integrate f by partial fractions. Check answers with computer or calculator assist. 7. fx) = x + 4 x + 5 7. fx) = x x 4 73. fx) = 74. fx) = 75. fx) = 76. fx) = 77. fx) = 78. fx) = 79. fx) = 8. fx) = x + 4 x + )x + ) xx ) x + )x + ) x + 4 x + )x + ) x x + )x + )) x + 4 x + )x + )x + 5) xx ) x + )x + )x + 3) x + 4 x + )x + )x ) xx ) x + )x + )x ) Special Methods. Integrate f by using the suggested u-substitution or method. Check answers with computer or calculator assist. 8. fx) = x + x + ), u = x +. 8. fx) = x + x ), u = x. 83. fx) = 84. fx) = x x + ) 3, u = x +. 3x x 3 + ), u = x3 +. 85. fx) = x3 + x, use long division. + 86. fx) = x4 + x, use long division. +