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Unformatted text preview: 1 ( x,t ) is indeed a solution. Similar procedure for u 2 ( x,t ) shows that it is also a solution (2.1.33) (6 pts ) The given ODE is ﬁrst order and linear. y + ay = beλt The integrating factor is μ = e R a · dt = e at Therefore ( ye at ) = e at ( beλt ) = ⇒ ye at = Z e at ( beλt ) · dt = b Z e ( aλ ) t · dt If a = λ , then we have ye at = b Z 1 · dt = bt + C = ⇒ y = bteat + Ceat Since both exponentials are negative ( a > 0), y → 0 as t → ∞ . ( evaluate the limit as t → ∞ ) If a 6 = λ ye at = b ± 1 aλ e ( aλ ) t ² + C 2 = ⇒ y = b ± 1 aλ e (λ ) t ² + Ceat Again, both exponentials are negative ( a > 0 and λ > 0 ), hence by similar reasoning, y → 0 as t → ∞ ) 3...
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 Spring '07
 COSTIN
 Math, Differential Equations, Equations, Partial Differential Equations, Trigraph, pts, Yeat, Dr. Huseyin Coskun, Total graded questions

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