Explanation Given: The differential equation is ( 4 y + 2 t − 5 ) d t + ( 6 y + 4 t − 1 ) d y = 0 and initial condition is y ( − 1 ) = 2 . Calculation: Consider the differential equation. ( 4 y + 2 t − 5 ) d t + ( 6 y + 4 t − 1 ) d y = 0 Compare the above equation with equation M ( x , y ) d x + N ( x , y ) d y = 0 . So, M = 4 y + 2 t − 5 and N = 6 y + 4 t − 1 . Now, d M d y = 4 and d N d t = 4 . That is, d M d y = d N d t . Therefore, the given equation is exact. Consider the equations below. d f d t = 4 y + 2 t − 5 … … ( 1 ) d f d y = 6 y + 4 t − 1 … … ( 2 ) Integrate equation (1) both sides

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 2.4, Problem 23E

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To solve: The given initial value problem.

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Chapter 2.4, Problem 22E

Chapter 2.4, Problem 24E

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Chapter 2 Solutions

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The given initial value problem.

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Chapter 2 Solutions