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Verify that the given differential equation is not exact. (x² + 2xy - y²) dx + (y² + 2xy = x²) dy = 0 - If the given DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has My = = 2x-2y Way to go! Nx = 2y-2x Since My and Nx are not equal, the equation is not exact. -2 Multiply the given differential equation by the integrating factor µ(x, y) = (x + y)¯² and verify that the new equation is exact. If the new DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has -4xy 3 Perfect! My = (x + y)³ -4xy NX = (x+y)3 Since My and Nx are - equal, the equation is exact. af (x² + 2xy + y² − 2y²). Integrate this partial derivative with respect to x, letting h(y) be an unknown function in y. Let = дх (x + y)² f(x, y) = Find the derivative of h(y). h' (y) = Solve. | x² + y² − c ( x + y) = 0 + h(y) Impressive work.

Verify that the given differential equation is not exact. (x² + 2xy - y²) dx + (y² + 2xy = x²) dy = 0 - If the given DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has My = = 2x-2y Way to go! Nx = 2y-2x Since My and Nx are not equal, the equation is not exact. -2 Multiply the given differential equation by the integrating factor µ(x, y) = (x + y)¯² and verify that the new equation is exact. If the new DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has -4xy 3 Perfect! My = (x + y)³ -4xy NX = (x+y)3 Since My and Nx are - equal, the equation is exact. af (x² + 2xy + y² − 2y²). Integrate this partial derivative with respect to x, letting h(y) be an unknown function in y. Let = дх (x + y)² f(x, y) = Find the derivative of h(y). h' (y) = Solve. | x² + y² − c ( x + y) = 0 + h(y) Impressive work.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Verify that the given differential equation is not exact. (x² + 2xy - y²) dx + (y² + 2xy = x²) dy = 0 - If the given DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has My = = 2x-2y Way to go! Nx = 2y-2x Since My and Nx are not equal, the equation is not exact. -2 Multiply the given differential equation by the integrating factor µ(x, y) = (x + y)¯² and verify that the new equation is exact. If the new DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has -4xy 3 Perfect! My = (x + y)³ -4xy NX = (x+y)3 Since My and Nx are - equal, the equation is exact. af (x² + 2xy + y² − 2y²). Integrate this partial derivative with respect to x, letting h(y) be an unknown function in y. Let = дх (x + y)² f(x, y) = Find the derivative of h(y). h' (y) = Solve. | x² + y² − c ( x + y) = 0 + h(y) Impressive work.
Transcribed Image Text:Verify that the given differential equation is not exact. (x² + 2xy - y²) dx + (y² + 2xy = x²) dy = 0 - If the given DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has My = = 2x-2y Way to go! Nx = 2y-2x Since My and Nx are not equal, the equation is not exact. -2 Multiply the given differential equation by the integrating factor µ(x, y) = (x + y)¯² and verify that the new equation is exact. If the new DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has -4xy 3 Perfect! My = (x + y)³ -4xy NX = (x+y)3 Since My and Nx are - equal, the equation is exact. af (x² + 2xy + y² − 2y²). Integrate this partial derivative with respect to x, letting h(y) be an unknown function in y. Let = дх (x + y)² f(x, y) = Find the derivative of h(y). h' (y) = Solve. | x² + y² − c ( x + y) = 0 + h(y) Impressive work.
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