Explanation Given: The given differential equation for initial value of y ( 1 ) = 2 is, x y 2 d y d x = y 3 − x 3 Procedure used: 1. Compare the given differential equation with the form M ( x , y ) d x + N ( x , y ) d y = 0 . 2. Check the homogeneity of the given differential equation. M ( t n x , t n y ) = t n M ( x , y ) N ( t n x , t n y ) = t n N ( x , y ) 3. If the given differential equation is homogeneous, then use the substitution y = u x . 4. Differentiate the equation y = u x with respect x and substitute the values in given differential equation. It will give another differential equation as P ( x ) d x = Q ( u ) d u . 5. Obtain the solution by integrating the equation P ( x ) d x = Q ( u ) d u . Calculation: Solve the given differential equation as, x y 2 d y d x = y 3 − x 3 ( x 3 − y 3 ) d x + x y 2 d y = 0 (1) Consider the below equation. y = u x (2) Differentiate the equation (2) with respect to x . d y d x = u + x d u d x d y = u d x + ( x d u d x ) ( d x ) d y = u d x + x d u (3) Substitute values of equation (2) and equation (3) in equation (1)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 2.5, Problem 11E

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To solve: The given differential equation using the given initial value.

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For each dif erential equation in Problems 1–21, find the general solutionby finding the homogeneous solution and a particular solution. Please DO NOT YOU THE PI method where 1/f(r) * x. Dont do that. Instead do this, assume for yp = to something, do the 1 and 2 derivative of it and then plug it in the equation to find the answer.

2.2. Solve the following initial value problems: a) t+2y=², y(2)=1. b) c) "+y=0, y(0)=1, (0) = -2. + 2y +2y=0, (0)=2, (0) = 1.

The solution of the following Initial Value Problem: у" +у%3D0 У (0) %3D2, у'(0) — 7 is:

Chapter 2 Solutions

Advanced Engineering Mathematics

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1.2 Find the general solution of dy -2x6 dr +y = y-4
Q. 67-75 Solve the initial value problems. dy 67. 2 – 7, y(2) = 0 dx 68. 10 - x, y(0) = -1 dx dy 69. - + x, > 0; y(2) = 1 dx 1-² dy 70. == 9x² - 4x + 5, y(-1) = 0 dx dy 71. = 3x-2/3, y(-1) = -5 dx dy 1 72. - y (4) = 0 dx 2Vx ds 73. = 1 + cost, s (0) = 4 dt 74. = cost + sint, s(7) = 1 -π sin 70, 7(0) = 0 1. / 126 + = = ds dt dr 75. d0 Evaluate the integrals in Exercises 13-36. 13. ·/₁ V3-2s ds 14. (2x + 1)³ dx Area In Exercises 37-42, fine x-axis. |37. y = -x − 2x, - 38. y = 3x²-3, -25 39. y=x²-3x² + 2 40. y = x² - 4x. -2 41. y = x¹/³, -1 s 42. y = x¹/3 - x, - 9. (a) Find the termi is (1, -2). (b) Find the initi point is (5, 0. 10. (a) Find the terr is (2, -1). (b) Find the ter point is (-2 11-12 Perform the and w. 11. u = 3i - k, v (a) w - v (c) -v - 2w (e) -8(v + w 12. u = (2,-1,3 (a) u - w

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The given differential equation using the given initial value.

Solution Summary: The author explains how to solve the given differential equation using the initial value of y(1)=2.

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