What is the solution for y^{''}+4y^'+4y=25cos(x) ?

The solution for y^{''}+4y^'+4y=25cos(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+4sin(x)+3cos(x)

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y^{''}+4y^'+4y=25cos(x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 25 cos (x)

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 25 cos (x)

Solve linear ODE:

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

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

n = 0 \sum \infty \frac{1}{3 ^{n} - n}

x^{y}

y = x^{2} 2 x - 7

\int (2 x^{2} + 1) e^{(x^{2})} d x

x \frac{d y}{d x} = \frac{1}{y ^{3}}

What is the solution for y^{''}+4y^'+4y=25cos(x) ?
The solution for y^{''}+4y^'+4y=25cos(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+4sin(x)+3cos(x)