Explanation Given information: The given equation is u ′ = u 2 + a x u − b , where a , b ≠ 0 that are real rational functions of x . Calculation: Consider that the Riccati equation has the form; d u d x = u 2 + a x u − b a , b ≠ 0 , that are real functions of x Introduce a new variable as shown below; z = 1 u − x This implies; u = 1 z + x Then, d u d x = − z ′ z 2 + 1 ................ ( 1 ) Since u = x is a solution to D E , Then, d u d x = x 2 + a − b = 1 Substitution of u = 1 z + x in D E yields; u ′ = ( 1 z + x ) 2 + a x ( 1 z + x ) − b − z ′ z 2 + 1 = ( x 2 + a = b ) + 1 z 2 + 2 x z + a x z z ′ = − 1 − 2 x z − a x z Therefore, z ′ + z ( 2 x + a x ) = − 1................ ( 2 ) As it can be seen, we have a linear first order ODE which can be solved by the method of integrating factor. Suppose that there is a differential equation of the form; d y d x + P ( x ) y = Q ( x ) Then integrating factor is u ( x ) = exp ( ∫ P ( x ) d x ) and general solution is as follows; u ( x ) y ( x ) = ∫ ( Q ( x ) ⋅ u ( x ) d x + C ) Here, integrating factor is; u ( x ) = exp ∫ ( 2 x + a x ) d x = e ( x 2 + a ln | x | ) = e x 2 x a General solution is as follows; u ( x ) z ( x ) = − ∫ e x 2 x a d x + C In order to solve ∫ e x 2 x a d x Put, v = x 2 This implies; d v = 2 x d x Substituting the values and simplify to get; ∫ e v v a 2 d v Integrand is an implicit it integral which cannot be solved by the use of conventional integration by parts method

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Differential Equations

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Chapter 14.1, Problem 60PS

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To find: All solutions of the given equation.

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Solution Summary: The author explains the general solution of the Riccati equation in u(x).

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Chapter 14 Solutions