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As a specific example we consider the non-homogeneous problem y" + 6y' + 8y = 4 sin(e²x)

As a specific example we consider the non-homogeneous problem y" + 6y' + 8y = 4 sin(e²x)

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 17EQ
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As a specific example we consider the non-homogeneous problem y" + 6y' + 8y = (1) The general solution of the homogeneous problem (called the complementary solution, Yc Ay₁ + By2 ) is given in terms of a pair of linearly independent solutions, Y₁, Y2. Here A and B are arbitrary constants. Find a fundamental set for y” + 6y' + 8y = 0 and enter your results as a comma separated list e^(-2x),e^(-4x) (2) Choose particula Y₁, Y2 in the fundamental set such that the W(x) is positive. Then W (x) 2e^(-6x) U 1 = If your W (x) is negative, then you need to switch your choise of y₁ and y2 and recalculate your W(x). The y₁= e^(-2x) ՂԱԶ √= With this appropriate order we are now ready to apply the method of variation of parameters. -Y₂(x) f(x) W(x) y = Yc + Yp and the y₂2 Y₁(x)f(x) W(x) S [21 dx -cos(e^(2x)) = dx = And combining these results we arrive at = S S e^(-4x) = 2e^(4x)sin(e^(2x)) dx 4 sin(e²x) dx = = = Ур (3) Finally, using A and B for the arbitrary constants in yc, the general solution can then be written as
Transcribed Image Text:As a specific example we consider the non-homogeneous problem y" + 6y' + 8y = (1) The general solution of the homogeneous problem (called the complementary solution, Yc Ay₁ + By2 ) is given in terms of a pair of linearly independent solutions, Y₁, Y2. Here A and B are arbitrary constants. Find a fundamental set for y” + 6y' + 8y = 0 and enter your results as a comma separated list e^(-2x),e^(-4x) (2) Choose particula Y₁, Y2 in the fundamental set such that the W(x) is positive. Then W (x) 2e^(-6x) U 1 = If your W (x) is negative, then you need to switch your choise of y₁ and y2 and recalculate your W(x). The y₁= e^(-2x) ՂԱԶ √= With this appropriate order we are now ready to apply the method of variation of parameters. -Y₂(x) f(x) W(x) y = Yc + Yp and the y₂2 Y₁(x)f(x) W(x) S [21 dx -cos(e^(2x)) = dx = And combining these results we arrive at = S S e^(-4x) = 2e^(4x)sin(e^(2x)) dx 4 sin(e²x) dx = = = Ур (3) Finally, using A and B for the arbitrary constants in yc, the general solution can then be written as
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