The unique solution to the initial value problem y y gt Y0

The unique solution to the initial value problem y\' + y = g(t), Y(0) = y_0, is y(t) = 5e^-4t + 9t - 2. Determine the constant y_0 and the function g(t). y_0 = help (numbers) g(t) = help (formulas)

Answer :

Given that y = 5e^-4t + 9t -2 then y\' = -20e^-4t + 9

Substitute y and y\' in y\' + y = g(t) we get

(- 20e^-4t + 9 )+( 5e^-4t + 9t -2 ) = g(t)

Therefore , g(t) = - 15e^-4t + 9t + 7

and y₀ = y(0) = 5e^-4(0) + 9(0) - 2 = 5 - 2 = 3

$The unique solution to the initial value problem y\' + y = g(t), Y(0) = y_0, is y(t) = 5e^-4t + 9t - 2. Determine the constant y_0 and the function g(t). y_0 =$

The unique solution to the initial value problem y y gt Y0

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