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The value of y as t→∞ for the following differential equation for an initial value of y(1) = 0 is(4t2+1)dydt+8yt−t=0 0.125 |
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Answer» The value of y as t→∞ for the following differential equation for an initial value of y(1) = 0 is (4t2+1)dydt+8yt−t=0
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