ENGINEERING MATHEMATICS - EXAMPLE 8.3 : Solve the first order differential equation : (Z + 1)(dy/dx) = xy in term of ln |y| = f(x). Z = (x)(x).
ENGINEERING MATHEMATICS - ANSWER 8.3 : Rearranging the equation (1/y)(dy) = [ x / (Z+1) ] (dx). Use the additional equation u = Z + 1, du = 2x dx, 0.5 du = x dx. Then the equation becomes (1/y)(dy) = (0.5)(1/u) du. Integrating both sides of the equation gives ln |y| = (0.5) ln |Z+1| + c where Z = square of x. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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ENGINEERING MATHEMATICS - EXAMPLE 8.3 : Solve the first order differential equation : (Z + 1)(dy/dx) = xy in term of ln |y| = f(x). Z = (x)(x).
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Question 72 - (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calory, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?
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