To determine The general solution of the ordinary differential equation y ′ = 6 ( y − 2.5 ) tanh 1.5 x . Explanation The given ordinary differential equation is y ′ = 6 ( y − 2.5 ) tanh 1.5 x . The ordinary differential equation is rewritten as follows. y ′ = 6 ( y − 2.5 ) tanh 1.5 x y ′ − y ⋅ 6 tanh ( 1.5 x ) + 15 tanh ( 1.5 x ) = 0 y ′ − y ⋅ 6 tanh ( 1.5 x ) = − 15 tanh ( 1.5 x ) The above differential equation is of the form y ′ + p y = r , where p = − 6 tanh ( 1.5 x ) and r = − 15 tanh ( 1.5 x ) . The general solution is given by, y ( x ) = e − h ( c + ∫ r e h d x ) ... ( 1 ) The integrating factor h is ∫ p d x and it is calculated as follows. ∫ p d x = − 6 ∫ tanh ( 1.5 x ) d x = − 6 ∫ sinh ( 1.5 x ) cosh ( 1.5 x ) d x = − 6 ∫ d t 1.5 t ( ∵ t = cosh ( 1.5 x ) , d t = sinh ( 1.5 x ) ⋅ ( 1.5 x ) d x ) = − 6 1.5 ∫ d t t = − 4 ln ( cosh x ) + c Substitute the integrating factor h in equation (1) and obtain the general solution as follows. y ( x ) = cosh 4 ( 1

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ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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Chapter 1.5, Problem 13P

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The general solution of the ordinary differential equation y′=6(y−2.5)tanh1.5x.

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