Explanation Given information: y ' − 1 2 ( 1 + t ) y = 1 + t , where t ≥ 0 Explanation: Step 1: Put the equation in the standard form: y ' − 1 2 ( 1 + t ) y = 1 + t Step 2: Find the integrating factor e A ( t ) Compare y ' − 1 2 ( 1 + t ) y = 1 + t and y ' + a ( t ) y = b ( t ) ⇒ a ( t ) = − 1 2 ( 1 + t ) , b ( t ) = 1 + t ⇒ A ( t ) = ∫ − 1 2 ( 1 + t ) d t = − 1 2 ln | 1 + t | = ln ( 1 + t ) − 1 2 Hence, the integrating factor is e A ( t ) = e ln ( 1 + t ) − 1 2 = ( 1 + t ) − 1 2 Step 3: Multiply both sides of the differential equation by the integrating factor ( 1 + t ) − 1 2 ( 1 + t ) − 1 2 y ' − ( 1 + t ) − 1 2 × 1 2 ( 1 + t ) y = ( 1 + t ) − 1 2

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ISBN: 9780137400096

Author: Larry Goldstein, David Lay

Publisher: PEARSON+

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The solution of the equation y' − 12(1+t)y = 1+t, where t≥0

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The solution of the equation y ' − 1 2 ( 1 + t ) y = 1 + t , where t ≥ 0

Solution Summary: The author explains how to determine the solution of the differential equation y

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The solution of the equation y' − 12(1+t)y = 1+t, where t≥0

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