m238sp06t2s - Mathematics 238 test two solutions Tuesday,...

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Mathematics 238 test two solutions Tuesday, May 16, 2006 1 . State the t -interval on which a solution to the following differential equation is guaranteed to exist: (4 - t 2 ) y 00 + 2 ty 0 + 6 y = ln t and y (1) = 0 and y 0 (1) = 0 Rewrite the equation so that the coefficient of the highest order derivative is equal to one: y 00 + ± 2 t 4 - t 2 ² y 0 + ± 6 4 - t 2 ² y = ln t 4 - t 2 The domain on which all the coefficients are continuous is 0 < t < 2 or t > 2. The solution to the initial condition involving t = 1 will exist inside this domain, so the solution will be guaranteed to exist when 0 < t < 2. 2 . Circle each linearly independent set of functions: ( a ) { f ( t ) = t 2 - 4 ,g ( t ) = t 3 - 4 t } linearly independent set since one function is not a constant multi- ple of the other ( b ) { f ( t ) = sin(2 t ) ,g ( t ) = sin( - 2 t ) } linearly dependent set since g ( t ) = ( - 1) f ( t ) ( c ) { F ( t ) = e 2 t ,G ( t ) = e - 2 t } linearly independent set since one function is not a constant multiple of the other
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m238sp06t2s - Mathematics 238 test two solutions Tuesday,...

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