What is the solution for y^{''}-7y^'+10y=-2cos(2t) ?

The solution for y^{''}-7y^'+10y=-2cos(2t) is y=c_{1}e^{5t}+c_{2}e^{2t}+7/58 sin(2t)-3/58 cos(2t)

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y^{''}-7y^'+10y=-2cos(2t)

y^{^{''}} - 7 y^{^{'}} + 10 y = - 2 cos (2 t)

y = c_{1} e^{5 t} + c_{2} e^{2 t} + \frac{7}{58} sin (2 t) - \frac{3}{58} cos (2 t)

Solution steps

y^{''} (t) - 7 y^{'} (t) + 10 y = - 2 cos (2 t)

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{2 t} + \frac{7}{58} sin (2 t) - \frac{3}{58} cos (2 t)

y = c_{1} e^{5 t} + c_{2} e^{2 t} + \frac{7}{58} sin (2 t) - \frac{3}{58} cos (2 t)

f (x) = 4 x - 1

\frac{d y}{d x} = 8 e^{x - y}

\int 3 v (v^{2} + 4)^{2} d v

\int x^{2} 5^{x} d x

\int \frac{10}{x} d x

What is the solution for y^{''}-7y^'+10y=-2cos(2t) ?
The solution for y^{''}-7y^'+10y=-2cos(2t) is y=c_{1}e^{5t}+c_{2}e^{2t}+7/58 sin(2t)-3/58 cos(2t)