To determine To solve: The initial value differential equation t y ′ + 2 y = t 2 − t + 1 , y ( 1 ) = 1 2 and t > 0 . Explanation Definition used: The first order linear equation is of the form d y d t + p ( t ) y = g ( t ) where p and g are function of the independent variable t . Formula used: The integrating factor μ ( t ) of the first order differential equation d y d t + p ( t ) y = g ( t ) is given as μ ( t ) = exp ∫ p ( t ) d t and the solution of the differential equation is given by y = 1 μ ( t ) [ ∫ μ ( s ) g ( s ) d s + c ] . Calculation: The given differential equation is t y ′ + 2 y = t 2 − t + 1 with the initial value y ( 1 ) = 1 2 and t > 0 . Rewrite the equation t y ′ + 2 y = t 2 − t + 1 as y ′ + 2 t y = t 2 − t + 1 t . This is possible since t > 0 . By comparing the equation y ′ + 2 t y = t 2 − t + 1 t with its standard form d y d t + p ( t ) y = g ( t ) , it is obtained that p ( t ) = 2 t and g ( t ) = t 2 − t + 1 t . Compute the integrating factor μ ( t ) of the differential equation y ′ + 2 t y = t 2 − t + 1 t as follows. μ ( t ) = exp ∫ p ( t ) d t = exp ∫ 2 t d t = exp ( 2 ln t ) = t 2 ( ∵ a ln b = ln b a and e ln c = c ) Now the solution of the differential equation y ′ + 2 t y = t 2 − t + 1 t can be obtained by substituting the integrating factor in the formula y = 1 μ ( t ) [ ∫ μ ( s ) g ( s ) d s + c ] as follows

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.1, Problem 15P

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To solve: The initial value differential equation ty′+2y=t2−t+1, y(1)=12 and t>0.

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The initial value differential equation t y ′ + 2 y = t 2 − t + 1 , y ( 1 ) = 1 2 and t &gt; 0 .

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To solve: The initial value differential equation ty′+2y=t2−t+1, y(1)=12 and t>0.

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The initial value differential equation t y ′ + 2 y = t 2 − t + 1 , y ( 1 ) = 1 2 and t > 0 .