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Unformatted text preview: . Answer: characteristic polynomial is , with roots , so . For , a particular solution should have the form Comparing terms, , , . For , a particular solution should have the form Final solution: 5. (6 pts.) Determine intervals where the solution for the following problem is sure to exist: Answer: the equation in standard form is , so the coefficient functions for and , and the function do not exist when . There should be a unique solution when or or . 6. (20 pts.) Use the variation of parameters method to solve the differential equation with initial values , and . Answer: characteristic polynomial is , with roots . , so The linear system is , . Adding first equation to second equation , the result is , so and ; using first equation, , so . Therefore ; now , and , so and . Final solution: ....
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This test prep was uploaded on 04/06/2008 for the course MATH 302 taught by Professor Goldberg during the Spring '08 term at Johns Hopkins.
 Spring '08
 GOLDBERG
 Math, Differential Equations, Equations

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