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Semmy [17]
3 years ago
11

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

Mathematics
1 answer:
MArishka [77]3 years ago
8 0
y'+\dfrac2ty=\dfrac{\cos t}{t^2}
t^2y'+2ty=\cos t
(t^2y)'=\cos t
t^2y=\sin t+C
y=\dfrac{\sin t+C}{t^2}
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