Explanation Calculation: The non-linear first-order differential equation is given by ( x 2 + y 2 − 5 ) d x = ( y + x y ) d y ( x 2 + y 2 − 5 ) d x − ( y + x y ) d y = 0 Compare the above differential equation with the general form of the exact differential equation. M = ( x 2 + y 2 − 5 ) N = − ( y + x y ) Partially differentiate M with respect to y . M y = 2 y Partially differentiate N with respect to x . N x = − y Consider the quotient. M y − N x N Substitute 2 y for M y , − y for N x , and − ( y + x y ) for N in the above equation. M y − N x N = 2 y − ( − y ) − ( y + x y ) = 3 y − y ( 1 + x ) = − 3 1 + x Find the integrating factor. I .F . = e ∫ ( M y − N x N ) d x Substitute − 3 1 + x for M y − N x N in the above equation. I F = e ∫ ( − 3 1 + x ) d x = e ln ( 1 + x ) − 3 = 1 ( 1 + x ) 3 Multiply the differential equation ( x 2 + y 2 − 5 ) d x − ( y + x y ) d y = 0 with 1 ( 1 + x ) 3 . ( x 2 + y 2 − 5 ) ( 1 + x ) 3 d x − ( y + x y ) ( 1 + x ) 3 d y = 0 \ The above equation is the exact differential. Comparing the above equation with the exact differential equation. M = ( x 2 + y 2 − 5 ) ( 1 + x ) 3 N = − ( y + x y ) ( 1 + x ) 3 N = − y ( 1 + x ) 2 Let, ∂ f ∂ y = N Substitute − y ( 1 + x ) 2 for N in the above equation. ∂ f ∂ y = − y ( 1 + x ) 2 ∂ f = − y ( 1 + x ) 2 ∂ y Integrate the above equation with respect to y . ∫ ∂ f = ∫ − y ( 1 + x ) 2 ∂ y f = − y 2 2 ( 1 + x ) 2 + h ( x ) (1) Here, h ( x ) is a function of x . Partially differentiate the above equation with respect to x . ∂ f ∂ x = y 2 ( 1 + x ) 3 + h ′ ( x ) Let, ∂ f ∂ y = M Substitute ( x 2 + y 2 − 5 ) ( 1 + x ) 3 for M in the above equation

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Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 2.4, Problem 38E

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The initial-value problem (x2+y2−5)dx=(y+xy)dy where y(0)=1 by finding an appropriate integrating factor.

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Chapter 2.4, Problem 37E

Chapter 2.4, Problem 39E

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Chapter 2 Solutions

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The initial-value problem ( x 2 + y 2 − 5 ) d x = ( y + x y ) d y where y ( 0 ) = 1 by finding an appropriate integrating factor.

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The initial-value problem (x2+y2−5)dx=(y+xy)dy where y(0)=1 by finding an appropriate integrating factor.

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Chapter 2 Solutions