Explanation The given differential equation is 1 t 2 ( 2 t 2 y cos t 2 − y sin t 2 ) d t + 1 t ( 2 t y + sin t 2 ) d y = 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 t 2 ( 2 t 2 y cos t 2 − y sin t 2 ) d t + 1 t ( 2 t y + sin t 2 ) d y = 0 Compare the given differential equation with M ( t , y ) d t + N ( t , y ) d y = 0 and obtain the value of M and N as shown below. Note that M ( t , y ) = 1 t 2 ( 2 t 2 y cos t 2 − y sin t 2 ) and N ( t , y ) = 1 t ( 2 t y + sin t 2 ) . Integrate M ( t , y ) with respect to t as follows. ∫ M ( t , y ) = ∫ 1 t 2 ( 2 t 2 y cos t 2 − y sin t 2 ) d t ∫ M ( t , y ) = ∫ 2 y cos t 2 d t − ∫ 1 t 2 y sin t 2 d t f ( t , y ) = − ( − y sin t 2 + 2 y t 2 cos t 2 t ) + 2 y t cos t 2 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 28E

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To solve: The differential equation 1t2(2t2ycost2−ysint2)dt+1t(2ty+sint2)dy=0.

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Chapter 2.5, Problem 27E

Chapter 2.5, Problem 29E

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Chapter 2 Solutions

Introductory Differential Equations

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The differential equation 1 t 2 ( 2 t 2 y cos t 2 − y sin t 2 ) d t + 1 t ( 2 t y + sin t 2 ) d y = 0 .

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To solve: The differential equation 1t2(2t2ycost2−ysint2)dt+1t(2ty+sint2)dy=0.

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Chapter 2 Solutions