Solve y (kx √√3) dx + (√3 − kx) dy O a. 2kx + 4√/31n|kx √3- O b. 2kx + 4√3ln|kx c. 2kx + 4√3 ln | kx - 2k In|y| √3 - 2k In ly √3+2k In y yz (kx + k sin (kz²) = C k sin (kz²) = C k sin (kz²) = C √3) cos (kz²) dz = 0

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter5: Polynomial And Rational Functions

Section: Chapter Questions

Problem 27PT: Find the unknown value. 27. y varies jointly with x and the cube root of 2. If when x=2 and...

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Question

Solve the differential equation using the separation of variables. Provide a complete solution. Use a handwritten and step-by-step calculation without the use of calculator. Explain the process and simplify your answer.

Solve
y (kx
√3) dx + (√√3 − kx) dy
kx) dy - yz (kx √√3) cos (kz²) dz = 0
+ k sin (kz²) = C
√√3| − 2k ln |y|
√3 - 2k In|y|
√3+2k In y
- k sin (kz²) = C
O a. 2kx + 4√3ln|kx
O b. 2kx + 4√/31n|kx
O c. 2kx +4√3ln|kx
-
k sin (kz²) = C

With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.

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