To determine To solve: The differential equation y ′ = ( 2 − e x ) ( 3 + 2 y ) , y ( 0 ) = 0 and compute the location in which its solution curve attains maximum value. Explanation Definition used: If it happens that M is a function x only and N is a function of y only such that M ( x ) d x + N ( y ) d y = 0 , then the differential equation d y d x = f ( x , y ) is separable. Calculation: The given differential equation is y ′ = ( 2 − e x ) ( 3 + 2 y ) , y ( 0 ) = 0 . In order to obtain the other solutions, rewrite the differential equation is y ′ = ( 2 − e x ) ( 3 + 2 y ) as follows. y ′ = ( 2 − e x ) ( 3 + 2 y ) d y d x = ( 2 − e x ) ( 3 + 2 y ) ( 3 + 2 y ) d y = ( 2 − e x ) d x ( 2 − e x ) d x − ( 3 + 2 y ) d y = 0 Note that M = 2 − e x , a function of x only and N = − ( 3 + 2 y ) , a function of y only. By above definition, the equation y ′ = ( 2 − e x ) ( 3 + 2 y ) is a separable differential equation. The separable differential equation y ′ = ( 2 − e x ) ( 3 + 2 y ) can be solved by solving the equation ( 2 − e x ) d x − ( 3 + 2 y ) d y = 0 as follows. ( 2 − e x ) d x − ( 3 + 2 y ) d y = 0 ∫ ( 2 − e x ) d x = ∫ ( 3 + 2 y ) d y 2 x − e x = 3 y + y 2 + c ( where c is constant ) Apply the initial condition y ( 0 ) = 0 in the above obtained answer to get the value of the constant c

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.2, Problem 24P

To determine

To solve: The differential equation y′=(2−ex)(3+2y), y(0)=0 and compute the location in which its solution curve attains maximum value.

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Chapter 2.2, Problem 23P

Chapter 2.2, Problem 25P

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