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

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

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

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

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t sin ( 2 t )}{4}

Solution steps

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

Solve linear ODE:

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t sin ( 2 t )}{4}

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t sin ( 2 t )}{4}

(sin (6 t) - 7)^{2}

\int \frac{cos ^{2} ( θ )}{sin ( θ )} d θ

f (x) = 2^{90}

x \to 5 + lim (x^{3} - c x)

f (x) = x + \frac{4}{x}

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