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

The solution for 4y^{''}+4y^'+y=cos(t) is y=c_{1}e^{-t/2}+c_{2}e^{-t/2}t+4/25 sin(t)-3/25 cos(t)

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4y^{''}+4y^'+y=cos(t)

4 y^{^{''}} + 4 y^{^{'}} + y = cos (t)

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{t}{2}} t + \frac{4}{25} sin (t) - \frac{3}{25} cos (t)

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + y = cos (t)

Solve linear ODE:

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{t}{2}} t + \frac{4}{25} sin (t) - \frac{3}{25} cos (t)

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{t}{2}} t + \frac{4}{25} sin (t) - \frac{3}{25} cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 x + 3

(d^{2} + d - 2) y = lo g_{6} (x)

(d^{2} + 5 d + 6) \cdot y = e^{x}

(d + 1)^{3} y = x^{2} e^{- x}

y^{^{''}} + 25 y = cos^{5} (x)

What is the solution for 4y^{''}+4y^'+y=cos(t) ?
The solution for 4y^{''}+4y^'+y=cos(t) is y=c_{1}e^{-t/2}+c_{2}e^{-t/2}t+4/25 sin(t)-3/25 cos(t)