H1_1030

# H1_1030 - 1 x,t is indeed a solution Similar procedure for...

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Math 415.01 - Dr. Huseyin Coskun 10:30 class - Solutions to graded problems - HW 1 January 12, 2011 The homework will be graded for 25 pts. Completion : 5 pts Total graded questions : 20 pts (7.3.30) (6 pts) We wish to show that λ = 0 of A is an eigenvalue ⇐⇒ A is singular. λ = 0 is an eigenvalue. ⇐⇒ | A - 0 · I | = 0 ⇐⇒ | A | = 0 ⇐⇒ A is singular. ( Note : The symbol ⇐⇒ stands for ‘if and only if’ and means that the statements imply each other in both directions ) (1.3.26) ( 8 pts ) We wish to show that the given functions are indeed solutions to the given PDE α 2 u xx = u tt . This can be checked by substituting for u and its derivatives. u 1 ( x,t ) = e - α 2 t sin x = u 1 x ( x,t ) = e - α 2 t cos x 1

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( diﬀ. wrt x ) = u 1 xx ( x,t ) = - e - α 2 t sin x (diﬀ. wrt x ) Similarly, u 1 t ( x,t ) = - α 2 e - α 2 t sin x ( diﬀ wrt t ) Substituting in the given PDE, LHS = α 2 ( - e - α 2 t sin x ) and RHS = - α 2 e - α 2 t sin x Since they are equal, u
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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 = be-λt The integrating factor is μ = e R a · dt = e at Therefore ( ye at ) = e at ( be-λt ) = ⇒ ye at = Z e at ( be-λt ) · dt = b Z e ( a-λ ) t · dt If a = λ , then we have ye at = b Z 1 · dt = bt + C = ⇒ y = bte-at + Ce-at Since both exponentials are negative ( a > 0), y → 0 as t → ∞ . ( evaluate the limit as t → ∞ ) If a 6 = λ ye at = b ± 1 a-λ e ( a-λ ) t ² + C 2 = ⇒ y = b ± 1 a-λ e (-λ ) t ² + Ce-at Again, both exponentials are negative ( a > 0 and λ > 0 ), hence by similar reasoning, y → 0 as t → ∞ ) 3...
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H1_1030 - 1 x,t is indeed a solution Similar procedure for...

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