m3354_f05_q1_sol

m3354_f05_q1_sol - h ( u, t ) so that h t = u 3 , h u = 3...

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Math 3354 Sec. 2, Quiz # 1 Name KEY 1. ( Pure Time Equation ) Solve the initial value problem u 0 = t 2 sin( t 3 ), u (0) = 0. u = Z t 2 sin( t 3 ) dt + C Set v = t 3 so that dv = 3 t 2 dt and our integral becomes u = Z t 2 sin( t 3 ) dt = 1 3 Z sin ( v ) dv = - 1 3 cos( t 3 ) + C 2. ( Exact Equation ) Given u 3 +3 tu 2 u 0 = 0. (a) show the equation is exact; (b) use this fact to find an implicit solution. f ( t, u ) = u 3 g ( t, u ) = 3 tu 2 f u ( t, u ) = 3 u 2 g t ( t, u ) = 3 u 2 Since f u = g t the equation is exact. Therefore, there exists a function
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Unformatted text preview: h ( u, t ) so that h t = u 3 , h u = 3 tu 2 . If we integrate the first equation with respect to t we get h = tu 3 + ` ( u ) . Thus we have 3 tu 2 = h u = 3 tu 2 + ` ( u ) which implies ` ( u ) = 0 so we take ` ( u ) = 0. Our answer is tu 3 = C ....
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This note was uploaded on 10/19/2011 for the course MATH 2254 taught by Professor Gilliam during the Fall '05 term at Texas Tech.

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