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 y ′ 1 y 2 = − e x is of the form f ( y ) y ′ = g ( x ) where, f ( y ) = 1 y 2 and g ( x ) = − e x . By the above theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ 1 y 2 d y = ∫ − e x d x ( y − 2 + 1 − 2 + 1 ) = − e x + C − 1 y = − e x + C Simplify further as, 1 = y ( e x + C ) y = 1 ( e x + C ) That is, the general solution of the differential equation y ′ = − y 2 e x is y = 1 ( e x + C )

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Chapter 9.2, Problem 34E

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To find: The general solution for the differential equation y′=−y2ex and its particular solution satisfying the initial condition y(0)=12.

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Chapter 9.2, Problem 33E

Chapter 9.2, Problem 35E

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The general solution for the differential equation y ′ = − y 2 e x and its particular solution satisfying the initial condition y ( 0 ) = 1 2 .

Solution Summary: The author explains the general and particular solution of the differential equation yprime.

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To find: The general solution for the differential equation y′=−y2ex and its particular solution satisfying the initial condition y(0)=12.

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