Program Explanation Given information: Differential equation is ( x 2 + 4 ) y ′ + 3 x y = x and initial condition is y ( 0 ) = 1 . Formula used: If differential equation is, y ′ + p ( x ) y = q ( x ) Then integrating factor is, I . F . = e ∫ p ( x ) d x Then general solution of the differential equation is, y ( I . F . ) = ∫ q ( x ) ( I . F . ) d x + C Calculation: Consider, given differential equation as, ( x 2 + 4 ) y ′ + 3 x y = x ....... (1) This equation can be rewritten as, y ′ + 3 x x 2 + 4 y = x x 2 + 4 ....... (2) Now, integrating factor of the equation is, I . F . = e ∫ ( 3 x x 2 + 4 ) d x = e 3 2 ln ( x 2 + 4 ) = e ln ( x 2 + 4 ) 3 2 = ( x 2 + 4 ) 3 2 Now, general solution of the equation is, y ( ( x 2 + 4 ) </

5th Edition

ISBN: 9780100802452

Author: Calvis

Publisher: YUZU

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Problems on Growing Functions

For a given set of functions f and g, the growth of functions states that the function g grows as fast as the function f. The growth rate of algorithms helps determine the algorithm's efficiency and compare its performance to other algorithms.

Question

Chapter 1.5, Problem 24P

Program Plan Intro

To find: The general solution of the differential equation (x2+4)y′+3xy=x . Also find particular solution when initial condition is y(0)=1

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Chapter 1.5, Problem 23P

Chapter 1.5, Problem 25P

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Chapter 1 Solutions

EBK DIFFERENTIAL EQUATIONS

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The general solution of the differential equation ( x 2 + 4 ) y ′ + 3 x y = x . Also find particular solution when initial condition is y ( 0 ) = 1

Solution Summary: The author explains that the general solution of the differential equation is (x2+4)yprime.

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To find: The general solution of the differential equation (x2+4)y′+3xy=x . Also find particular solution when initial condition is y(0)=1

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Chapter 1 Solutions