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dy ху = x³e-*sin x dx + y² 5.

dy ху = x³e-*sin x dx + y² 5.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Item 5. Determine what type of differential equation and its solution. 

TEST YOUR UNDERSTANDING! Match the following differential equations and their general or particular solution. Show your complete step-by- step solution. DIFFERENTIAL EQUATION SOLUTION (2y³ + 4) dx = (6² – 2xy') dy )dx = (6²- 2xy:) dy 1. а. y? = e-* с у? 2. yy" + (y')? = yy' y(0) = 1, y'(0) =; b. x2 y 3 3. (2ху)dx + (y? + х?)dy %3D 0 C %3 (2х — Зу)^3 (х + у)^2 С. 9 y + 2 4. (V² + y² -)dx + G + y² – 1)dy EVx? + y² – 1) dy C =In (sec(tan-1 (6) +) d. у — 1 = 0 1 - csc (tan *(5))- 3(y– 1) -1 3(у — 1) | dy 2 - x³e-*sin x +y² = [C(x – C2)j½ 5. ху е. s = - dx dy 2x = (3y + x) dx C = x²y² + 4x – 6y f. dx 7. y2. + 2xy = sin y ,x(n) = 1 g. C = (x2 + y?)z – xy dy 3 t -2xcsc 2y – 2x³cos 2y 2 In sin +C dx 8. h. S = dy sec 2y t cos 9. (y³ – 3y + 2)/9 + x²dx + x²dy = 0 i. C = s?t – tet + et 10. (x² – 1)y" + 2y' – (x – 1)(x + 1)² = 0 j. у2 3D —х?е *(sin x + cos x )+ Сх2 2y2 + 6у — 1 :%==V16 – x² – 8sin-1Ó C 4 11. x²dx – /16 – x²(3y + y²) sin 2y k. 2у + 3 cos 2y + 4. sin 2y п2 — 1 —сos y 12. 2ss" + (s')² = 0 I. X = y2 ds 1 1 13. sin 4y + (C – 4y)sin² (2y) s tan =csC m. dt 2 2 2 x2 x3 14. 2ste-tds + (s²e-t – t)dt = 0 + (C, – 1)x + (2C, – 1) In |x – 1| + C2 6. n. y = e. 6.
Transcribed Image Text:TEST YOUR UNDERSTANDING! Match the following differential equations and their general or particular solution. Show your complete step-by- step solution. DIFFERENTIAL EQUATION SOLUTION (2y³ + 4) dx = (6² – 2xy') dy )dx = (6²- 2xy:) dy 1. а. y? = e-* с у? 2. yy" + (y')? = yy' y(0) = 1, y'(0) =; b. x2 y 3 3. (2ху)dx + (y? + х?)dy %3D 0 C %3 (2х — Зу)^3 (х + у)^2 С. 9 y + 2 4. (V² + y² -)dx + G + y² – 1)dy EVx? + y² – 1) dy C =In (sec(tan-1 (6) +) d. у — 1 = 0 1 - csc (tan *(5))- 3(y– 1) -1 3(у — 1) | dy 2 - x³e-*sin x +y² = [C(x – C2)j½ 5. ху е. s = - dx dy 2x = (3y + x) dx C = x²y² + 4x – 6y f. dx 7. y2. + 2xy = sin y ,x(n) = 1 g. C = (x2 + y?)z – xy dy 3 t -2xcsc 2y – 2x³cos 2y 2 In sin +C dx 8. h. S = dy sec 2y t cos 9. (y³ – 3y + 2)/9 + x²dx + x²dy = 0 i. C = s?t – tet + et 10. (x² – 1)y" + 2y' – (x – 1)(x + 1)² = 0 j. у2 3D —х?е *(sin x + cos x )+ Сх2 2y2 + 6у — 1 :%==V16 – x² – 8sin-1Ó C 4 11. x²dx – /16 – x²(3y + y²) sin 2y k. 2у + 3 cos 2y + 4. sin 2y п2 — 1 —сos y 12. 2ss" + (s')² = 0 I. X = y2 ds 1 1 13. sin 4y + (C – 4y)sin² (2y) s tan =csC m. dt 2 2 2 x2 x3 14. 2ste-tds + (s²e-t – t)dt = 0 + (C, – 1)x + (2C, – 1) In |x – 1| + C2 6. n. y = e. 6.
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