Explanation Given: The differential equation is ( 1 + 5 t − y ) d t − ( t + 2 y ) d y = 0 and the initial condition is y ( 0 ) = 0 . Procedure used: Solution to a differential equation : (1) Assume M ( t , y ) = ∂ f / ∂ t ( t , y ) and N ( t , y ) = ∂ f / ∂ y ( t , y ) . (2) Integrate M ( t , y ) with respect to t . (For arbitrary constant, add an arbitrary function of y , g ( y ) .) (3) Differentiate the result in step 2 with respect to y and set the result equal to N ( t , y ) . Solve for g ′ ( y ) . (4) Integrate g ′ ( y ) with respect to y to obtain an expression for g ( y ) . (5) Substitute g ( y ) into the result obtained in step 2 for f ( t , y ) . (6) A general solution f ( t , y ) = C , where C is an arbitrary constant. (7) Apply the initial condition if given. Calculation: Consider the given differential equation. ( 1 + 5 t − y ) d t − ( t + 2 y ) d y = 0 Here, M ( t , y ) = ( 1 + 5 t − y ) and N ( t , y ) = − ( t + 2 y ) . Integrate M ( t , y ) with respect to t ∫ M ( t , y ) = ∫ ( 1 + 5 t − y ) d t f ( t , y ) = t + 5 ( t 2 2 ) − t y + g ( y ) Differentiate f ( t , y ) with respect to y

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 2.5, Problem 34E

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The solution of differential equation (1+5t−y)dt−(t+2y)dy=0.

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Chapter 2.5, Problem 33E

Chapter 2.5, Problem 35E

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Chapter 2 Solutions

Introductory Differential Equations

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The solution of differential equation ( 1 + 5 t − y ) d t − ( t + 2 y ) d y = 0 .

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The solution of differential equation (1+5t−y)dt−(t+2y)dy=0.

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Chapter 2 Solutions