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 2 y ′ = x 2 is of the form f ( y ) y ′ = g ( x ) where, f ( y ) = y 2 and g ( x ) = x 2 . By the above mentioned theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ y 2 d y = ∫ x 2 d x y 2 + 1 2 + 1 = x 2 + 1 2 + 1 + C y 3 = x 3 + 3 C y = ( x 3 + 3 C ) 1 3 That is, the general solution of the differential equation y ′ = x 2 y 2 is y = ( x 3 + 3 C ) 1 3 . Apply the initial condition y ( 0 ) = 2 in y = ( x 3 + 3 C ) 1 3 to find the particular solution

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Author: Barnett

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Chapter 9.2, Problem 32E

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To find: The general solution for the differential equation y′=x2y2 and its particular solution satisfying the initial condition y(0)=2.

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Chapter 9.2, Problem 31E

Chapter 9.2, Problem 33E

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

Solution Summary: The author explains the general and particular solution of the differential equation yprime, which is given implicitly by the equation.

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To find: The general solution for the differential equation y′=x2y2 and its particular solution satisfying the initial condition y(0)=2.

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