What is the solution for x^{''}-4x^'+5x=cos(t) ?

The solution for x^{''}-4x^'+5x=cos(t) is x=e^{2t}(c_{1}cos(t)+c_{2}sin(t))-1/8 sin(t)+1/8 cos(t)

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x^{''}-4x^'+5x=cos(t)

x^{^{''}} - 4 x^{^{'}} + 5 x = cos (t)

x = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{1}{8} sin (t) + \frac{1}{8} cos (t)

Solution steps

x^{''} (t) - 4 x^{'} (t) + 5 x = cos (t)

Solve linear ODE:

x = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{1}{8} sin (t) + \frac{1}{8} cos (t)

x = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{1}{8} sin (t) + \frac{1}{8} cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{2 x}, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{'''}} + 4 y^{^{'}} = t e^{- 2 t} + 3 cos (2 t)

y^{^{''}} - y^{^{'}} = 3 e^{t}

y^{^{'''}} - 3 \cdot y^{^{''}} + 2 \cdot y^{^{'}} = - 3 \cdot x^{2} + 6

(d^{2} - 1) y = \frac{2}{1 + e ^{x}}

What is the solution for x^{''}-4x^'+5x=cos(t) ?
The solution for x^{''}-4x^'+5x=cos(t) is x=e^{2t}(c_{1}cos(t)+c_{2}sin(t))-1/8 sin(t)+1/8 cos(t)