To determine To solve: The differential equation ( t − 1 ) 3 d s d t + 4 ( t − 1 ) 2 s = t + 1 , t > 1 . Explanation Procedure used: To solve the linear equation y ′ + P ( x ) y = Q ( x ) , multiply both sides by the integrating factor v ( x ) = e ∫ P ( x ) d x and integrate on both sides. The equation can also be solved directly by using the formula y = 1 v ( x ) ∫ v ( x ) Q ( x ) d x . Calculation: Rewrite the differential equation ( t − 1 ) 3 d s d t + 4 ( t − 1 ) 2 s = t + 1 , t > 1 in the standard form as follows. ( t − 1 ) 3 d s d t + 4 ( t − 1 ) 2 s = t + 1 d s d t + 4 t − 1 s = t + 1 ( t − 1 ) 3 Compare the above equation with the standard form d y d x + P ( x ) y = Q ( x ) . Here, P ( t ) = 4 t − 1 and Q ( t ) = t + 1 ( t − 1 ) 3 . Evaluate the value of ∫ P ( t ) d t as follows. ∫ P ( t ) d t = ∫ 4 t − 1 d t = 4 ln | t − 1 | = ln ( t − 1 ) 4 ( ∵ t > 1 ) Evaluate the integrating factor as follows. v ( t ) = e ∫ P ( t ) d t = e ln ( t − 1 ) 4 = ( t − 1 ) 4 Thus, the integrating factor is ( t − 1 ) 4

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ISBN: 9780134439099

Author: Hass, Joel., Heil, Christopher , WEIR, Maurice D.

Publisher: Pearson,

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Chapter 9.2, Problem 11E

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To solve: The differential equation (t−1)3dsdt+4(t−1)2s=t+1,t>1.

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Chapter 9.2, Problem 10E

Chapter 9.2, Problem 12E

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Chapter 9 Solutions

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The differential equation ( t − 1 ) 3 d s d t + 4 ( t − 1 ) 2 s = t + 1 , t &gt; 1 .

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To solve: The differential equation (t−1)3dsdt+4(t−1)2s=t+1,t>1.

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The differential equation ( t − 1 ) 3 d s d t + 4 ( t − 1 ) 2 s = t + 1 , t > 1 .