Ch 9, 10 Differential Equations Notes

Ch 9, 10 Differential Equations Notes - 1.a Chap 9,10...

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1 Chap. 9,10 Differential Equations Differential equations are equations that contain derivatives. To solve the equation we must find the original function for which we are given the derivative. The original function is called the anti derivative. Notation: If f(x) is the derivative of F(x) i.e. If F / (x) = f(x) then F(x) is the anti derivative of f(x) eg. F(x) = x 2 + 11 F(x) = x 2 3 F(x) = x 2 + C F / (x) = 2x F / (x) = 2x F / (x) = 2x i.e. if F / (x) = f(x) = 2x then the anti derivative is F(x) = x 2 + C Note that we must always add the arbitrary constant, C, to every anti derivative because when differentiating the derivative of all constants is 0. derivative anti derivative F / (x) = 0 F(x) = C F / (x) = 1 F(x) = x + C F / (x) = 7 F(x) = 7x + C F / (x) = 2x F(x) = x 2 + C F / (x) = 6x F(x) = 3x 2 + C F / (x) = x n F(x) = 1 1 1 n x C n , n 1 F / (x) = 1 x F(x) = ln x C F / (x) = cosx F(x) = sinx + C F / (x) = sin12x F(x) = 1 cos12 12 x C F / (x) = kx e F(x) = 1 kx e C k Note k and e are constants F / (x) = kx a F(x) = 1 1 ln kx a C a k F / (x) = 2 3 3 8 x x F(x) = 3 ln 8 x C F / (x) = sin cos x x e F(x) = sin x e C F / (x) = 5 2 3 3 8 x x F(x) = 6 3 1 8 6 x C F / (x) = sec 2 (5x) + 4x 2 F(x) = 2 1 tan(5 ) 2 2 5 x x x C The actual value of the arbitrary constant, C, can often be found if we have extra information. 1.a
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2 1. A method that works: If F / (x) = 7 2 3 4 2 8 3 x x Assume F (x) = 11 3 4 8 k x So F / (x) = 7 7 3 2 2 3 4 4 11 33 8 (3 ) 8 4 4 k x x kx x Then 2 33 8 , 3 4 99 k so k and F (x) = 11 3 4 8 8 99 x C 2. The slope of a curve is given by the function 2 2 4 dy m x x dx . What is the equation of the particular function with this slope whose graph is on the point (3,5)? / 2 ( ) 2 4 dy F x m x x dx 3 2 1 ( ) 4 3 F x x x x C but (3, 5) is on the curve so F(3) = 5 or 3 2 1 (3) 5 3 3 4 3 3 F C 5 = 9 9 + 12 + C so C = 7 and 3 2 1 ( ) 4 7 3 F x x x x 3. What are the velocity and displacement functions if the acceleration is given by a = 3t 2 2; the initial velocity is 4 m/s and the displacement is 20m at 2 seconds. ( at t = 0, v = 4) 2 3 2 dv a t dt 3 2 then v t t C at t = 0, v = 4 so 4 = 0 3 2(0) + C 4 = C and v = t 3 2t + 4 3 2 4 ds v t t dt 4 2 1 4 4 s t t t K but s = 20 at t = 2 4 2 1 20 (2) (2) 4(2) 4 K 20 = 4 4 + 8 + K, so K = 12 and 4 2 1 4 12 4 s t t t Read p 400, 403 407, 409 411. Do p408 # (1 6) a, b ; 7 (all) p411 #(1 4) a and b 1.b
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3 More difficult anti derivatives For a simple polynomial it is easy to find the anti derivative using F / (x) = x n means F(x) = 1 1 1 n x C n , n 1 so if F / (x) = x 5 + 7x 3 3x + 6 then F(x) = 6 4 2 1 7 3 6 6 4 2 x x x x C but if the differentiation involves the chain rule as well we must anticipate the term generated by the chain rule as it will not be in the original function. 1. if y = (7x 3 + 3) 4 then y / = 4(7x 3 + 3) 3 (21x 2 ) y / = 84x 2 (7x 3 + 3) 3 so if F / (x) = x 2 (7x 3 + 3) 3 or if F / (x) = 13x 2 (7x 3 + 3) 3 then F(x) = 4 3 1 7 3 84 x C then F(x) = 4 3 13 7 3 84 x C Note that the x 2 term generated by the chain rule was not present in the original function.
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