Use variation of parameters to find a general solution to the differential equation given that the functions y₁ and y₂ are linearly independent solutions to the corresponding homogeneous equation for t > 0. ty"' + (5t-1)y' - 5y = 2t²e-5t. A general solution is y(t)= - 5t y₁ = 5t-1₁ y₂ = e

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