What is the solution for y^{''}+4y^'+4y=5cos(4x) ?

The solution for y^{''}+4y^'+4y=5cos(4x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+1/5 sin(4x)-3/20 cos(4x)

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y^{''}+4y^'+4y=5cos(4x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 cos (4 x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{5} sin (4 x) - \frac{3}{20} cos (4 x)

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 5 cos (4 x)

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{5} sin (4 x) - \frac{3}{20} cos (4 x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{5} sin (4 x) - \frac{3}{20} cos (4 x)

f (x) = 5 x^{\frac{2}{3}} - 2 x^{3}

\frac{1}{( s - 2 ) ^{6}}

\frac{d}{d x} (x^{2} + 2 x + 5)

x \to 0 + lim (\frac{1}{x ^{5}})

\int \frac{x}{1 - x ^{4}} d x

What is the solution for y^{''}+4y^'+4y=5cos(4x) ?
The solution for y^{''}+4y^'+4y=5cos(4x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+1/5 sin(4x)-3/20 cos(4x)