What is the solution for y^{''}-2y^'+5y=2cos^2(t) ?

The solution for y^{''}-2y^'+5y=2cos^2(t) is

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y^{''}-2y^'+5y=2cos^2(t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 cos^{2} (t)

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{t} cos (2 t) \cdot \int - \frac{cos ^{2} ( t ) sin ( 2 t )}{e ^{t}} d t + e^{t} sin (2 t) \cdot \int \frac{cos ^{2} ( t ) cos ( 2 t )}{e ^{t}} d t

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 2 cos^{2} (t)

Solve linear ODE:

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{t} cos (2 t) \cdot \int - \frac{cos ^{2} ( t ) sin ( 2 t )}{e ^{t}} d t + e^{t} sin (2 t) \cdot \int \frac{cos ^{2} ( t ) cos ( 2 t )}{e ^{t}} d t

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{t} cos (2 t) \cdot \int - \frac{cos ^{2} ( t ) sin ( 2 t )}{e ^{t}} d t + e^{t} sin (2 t) \cdot \int \frac{cos ^{2} ( t ) cos ( 2 t )}{e ^{t}} d t

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 e^{- t}, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 7 y^{^{'}} + 10 y = 9 cos (t) + 7 sin (t)

y^{^{''}} - y = cosh (x), y (0) = 2, y^{^{'}} (0) = 12

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{4 t}

y^{^{''}} + 2 y^{^{'}} + 2 y = 3, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+5y=2cos^2(t) ?
The solution for y^{''}-2y^'+5y=2cos^2(t) is