Explanation Theorem used: “If the functions p, q and g are continuous on an open interval I, and if the functions y 1 and y 2 are a fundamental set of solutions of the homogeneous equation of y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = 0 then a particular solution can be obtained from Y ( t ) = − y 1 ( t ) ∫ t 0 t y 2 ( s ) g ( s ) W ( y 1 , y 2 ) ( s ) d s + y 2 ( t ) ∫ t 0 t y 1 ( s ) g ( s ) W ( y 1 , y 2 ) ( s ) d s , where t 0 is any conveniently chosen point in I . The general solution is y = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Y ( t ) .” Calculation: The given differential equation is ( 1 − t ) y ″ + t y ′ − y = 2 ( t − 1 ) 2 e − t , 0 < t < 1 . Rewrite the above differential equation as y ″ + t ( 1 − t ) y ′ − 1 ( 1 − t ) y = 2 ( 1 − t ) e − t . Compare the obtained equation y ″ + t ( 1 − t ) y ′ − 1 ( 1 − t ) y = 2 ( 1 − t ) e − t with its standard differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) and get p ( t ) = t 1 − t and g ( t ) = 2 ( 1 − t ) e − t . Given that, y 1 ( t ) = e t and y 2 ( t ) = t . Substitute y 1 ( t ) = e t in ( 1 − t ) y ″ + t y ′ − y as follows. ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) d 2 d x 2 ( e t ) + t d d t ( e t ) − ( e t ) = ( 1 − t ) e t + t e t − e t = e t ( 1 − t + t − 1 ) = 0 Thus, the function y 1 ( t ) = e t is a solution of the homogeneous equation of the differential equation ( 1 − t ) y ″ + t y ′ − y = 2 ( t − 1 ) 2 e − t , 0 < t < 1 . Similarly, substitute y 2 ( t ) = t in ( 1 − t ) y ″ + t y ′ − y as follows. ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) d 2 d x 2 ( t ) + t d d t ( t ) − ( t ) = 0 + t − t = 0 + 0 = 0 Thus, the function y 2 ( t ) = t is a solution of the homogeneous equation of the differential equation ( 1 − t ) y ″ + t y ′ − y = 2 ( t − 1 ) 2 e − t , 0 < t < 1 . Thus, it is verified that the functions y 1 ( t ) = e t and y 2 ( t ) = t are the fundamental set of solutions of the homogeneous equation. Find the wronskian of the functions y 1 ( t ) = e t and y 2 ( t ) = t

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3.6, Problem 16P

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To verify: Whether the functions y1(t)=et and y2(t)=t satisfies the corresponding homogeneous equation of the differential equation (1−t)y″+ty′−y=2(t−1)2e−t, 0<t<1 and compute the particular solution.

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Chapter 3.6, Problem 15P

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To verify: Whether the functions y 1 ( t ) = e t and y 2 ( t ) = t satisfies the corresponding homogeneous equation of the differential equation ( 1 − t ) y ″ + t y ′ − y = 2 ( t − 1 ) 2 e − t , 0 &lt; t &lt; 1 and compute the particular solution.

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To verify: Whether the functions y1(t)=et and y2(t)=t satisfies the corresponding homogeneous equation of the differential equation (1−t)y″+ty′−y=2(t−1)2e−t, 0<t<1 and compute the particular solution.

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To verify: Whether the functions y 1 ( t ) = e t and y 2 ( t ) = t satisfies the corresponding homogeneous equation of the differential equation ( 1 − t ) y ″ + t y ′ − y = 2 ( t − 1 ) 2 e − t , 0 < t < 1 and compute the particular solution.