Verify that y_1 = e^2t and y_2 = te^2t are solutions of the homogeneous DE y\" -

Question

Verify that y_1 = e^2t and y_2 = te^2t are solutions of the homogeneous DE y" - 4y' + 4y = 0. Verify that their linear combination y = c_1 e^2t + c_2 te^2t where c_1 and c_2 are constants is also a solution of the homogeneous DE. Here, y' = dy/dt. Verify that y_1 = e^t/6 + e^-2t and y_2 = e^t/6 + e^-t are solutions of the nonhomogeneous DE y" + 3y' + 2y = e^t. Verify that their linear combination y = c_1 e^2t + c_2 te^2t where c_1 and c_2 are constants is also a solution of the homogeneous DE. Here, y' = dy/dt. Find the general solution of the DE y" + 7y' - 8y = 0. Then, find the specific solution to the DE with y(0) = 2 and y'(0) = 1.

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