hwk_06_sols

# hwk_06_sols - Solutions to Homework 6 Introduction to...

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Solutions to Homework 6, Introduction to Differential Equations, 3450:335-003, Dr. Montero, Spring 2009 Problem 1. Find a particular solution to y ′′ + 6 y + 9 y = - 25 e 3 t 2(1 + t 2 ) . Solution: First we find the solutions to the homogeneous equation y ′′ + 6 y + 9 y = 0 . The polynomial associated to this equation is λ 2 + 6 λ + 9 = ( λ + 3) 2 . The root is 3 with multiplicity 2. The solutions to the homogeneous problem are y 1 = e 3 t and y 2 = te 3 t . We will need the Wronskian for these two functions. We compute this to obtain W ( t ) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle y 1 ( t ) y 2 ( t ) y 1 ( t ) y 2 ( t ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle e 3 t te 3 t - 3 e 3 t e 3 t - 3 te 3 t vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = e 6 t . We seek a particular solution of the form y p ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) . For this to work we need u 1 ( t ) = - te 3 t × ( - 25) e 3 t 2(1 + t 2 ) e 6 t = 25 t 2(1 + t 2 ) = 25 4 2 t 1 + t 2 . We integrate this to obtain u 1 ( t ) = 25 4 ln(1 + t 2 ) . We also need u 2 = e 3 t × 25 e 3 t 2(1 + t 2 ) e 6 t = 25 2 1 1 + t 2 . We find that u 2 = 25 2 arctan( t ) .

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The particular solution we seek is y p ( t ) = 25 4 ln(1 + t 2 ) e 3 t + 25 2 arctan( t ) te 3 t .
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