Explanation Given information: The provided first order linear initial value problem is x d y d x − 2 y = x 3 . The above equation passes through ( 2 , − 1 ) and the graphical solution is, Concept used: The first order linear initial value problem is of the form, d y d x + p ( x ) y = q ( x ) The general solution is given by, y = 1 I ( x ) [ ∫ I ( x ) q ( x ) d x + C ] ; where I ( x ) is the integrating factor and equals to I ( x ) = e ∫ p ( x ) d x . Consider the first order linear initial value problem x d y d x − 2 y = x 3 . The provided equation can be written as, x d y d x − 2 y = x 3 d y d x − 2 y x = x 2 The integrating factor is, I ( x ) = e ∫ ( − 2 x ) d x = e − 2 ln x = 1 x 2

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 14.1, Problem 11PS

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The solution of the first order linear initial value problem xdydx−2y=x3 and compare the result with the provided graphical solution.

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Chapter 14.1, Problem 10PS

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+ I/ MI * 00 Energy consumption in a particular country in quadrillion BTUS can be modeled by C(x) = - 0.014x + 1.295x+ 68,958, where x is the number of years after 1970. a. One solution to the equation 89.258 = - 0.014x+ 1.295x+68.958 is x = 20. What does this mean? b. Graphically verify that x 20 is a solution to 89.258 = - 0.014x+1.295x + 68.958. c. To find when after 2020 the energy consumption in that country will be 89.258 quadrillion BTUS according to the model, is there a need to find the second solution to this equation? Why or why not? a. Choose the correct answer below. O A. In 1990, consumption was 68.958 quadrillion BTUS. O B. In 2020, consumption was 68.958 quadrillion BTUS. O C. In 1990, consumption was 89.258 quadrillion BTUS. O D. In 2020, consumption was 89.258 quadrillion BTUS. Click to select your answer and then click Check Answer. parts 2. remaining Clear All Check Answer P Pearson &17 PM ere to search 近 1202/LL/Z PrtSc F11 F12 F10 sup F7 F8 & Backspace 23 3. i 5.…

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Ch. 14.1 - Prob. 1PS Ch. 14.1 - Prob. 2PS Ch. 14.1 - Prob. 3PS Ch. 14.1 - Prob. 4PS Ch. 14.1 - Prob. 5PS Ch. 14.1 - Prob. 6PS Ch. 14.1 - Prob. 7PS Ch. 14.1 - Prob. 8PS Ch. 14.1 - Prob. 9PS Ch. 14.1 - Prob. 10PS

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The solution of the first order linear initial value problem x d y d x − 2 y = x 3 and compare the result with the provided graphical solution.

Solution Summary: The author compares the solution of the first order linear initial value problem xdy dx to the provided graphical solution.

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The solution of the first order linear initial value problem xdydx−2y=x3 and compare the result with the provided graphical solution.

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Chapter 14 Solutions