Explanation Theorem used: Separation of Variables: The solution of the differential equation, f ( y ) y ′ = g ( x ) is given implicitly by the equation, ∫ f ( y ) d y = ∫ g ( x ) d x . Calculation: The given differential equation can be rewritten as, x y ′ = x y + 2 y x y ′ = y ( x + 2 ) 1 y y ′ = ( x + 2 ) x y − 1 2 y ′ = 1 + 2 x That is, y − 1 2 y ′ = 1 + 2 x (1) The given differential equation y − 1 2 y ′ = 1 + 2 x is in the form of f ( y ) y ′ = g ( x ) where, f ( y ) = y − 1 2 and g ( x ) = 1 + 2 x . By the above theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ y − 1 2 d y = ∫ ( 1 + 2 x ) d x y − 1 2 + 1 − 1 2 + 1 = x + 2 ln x + C y 1 2 ( 1 2 ) = x + ln x 2 + C Simplify further as, 2 y 1 2 = x + ln x 2 + C ( 2 y 1 2 ) 2 = ( x + ln ( x 2 ) + C ) 2 4 y =

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Chapter 9.2, Problem 49E

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To find: The particular solution of the differential equation xy′=xy+2y satisfying the initial condition. y(1)=4.

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Chapter 9.2, Problem 48E

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The particular solution of the differential equation x y ′ = x y + 2 y satisfying the initial condition. y ( 1 ) = 4 .

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To find: The particular solution of the differential equation xy′=xy+2y satisfying the initial condition. y(1)=4.

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