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13-10-2017
Mathematics

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Find the solution of the given initial value problem y'+2/ty=cost/t^2

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LammettHash LammettHash

13-10-2017

[tex]y'+\dfrac2ty=\dfrac{\cos t}{t^2}[/tex]
[tex]t^2y'+2ty=\cos t[/tex]
[tex](t^2y)'=\cos t[/tex]
[tex]t^2y=\sin t+C[/tex]
[tex]y=\dfrac{\sin t+C}{t^2}[/tex]

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