Solve the given initial-value problem. (3y 2 - t 2 /y 5 ) dy/dt + t/2y
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Solve the given initial-value problem.
(3y2 - t2/y5) – dy/dt + t/2y4 = 0, y(1) = 1
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Let M t2y 4 and N 3y 2 t 2 y 5 So that M y 2ty 5 ...View the full answer
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Related Book For
A First Course in Differential Equations with Modeling Applications
ISBN: 978-1305965720
11th edition
Authors: Dennis G. Zill
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