What is the solution for y^{''}+20y^'+100y=240cos(10x) ?

The solution for y^{''}+20y^'+100y=240cos(10x) is y=c_{1}e^{-10x}+c_{2}xe^{-10x}+6/5 sin(10x)

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y^{''}+20y^'+100y=240cos(10x)

y^{^{''}} + 20 y^{^{'}} + 100 y = 240 cos (10 x)

y = c_{1} e^{- 10 x} + c_{2} x e^{- 10 x} + \frac{6}{5} sin (10 x)

Solution steps

y^{''} (x) + 20 y^{'} (x) + 100 y = 240 cos (10 x)

Solve linear ODE:

y = c_{1} e^{- 10 x} + c_{2} x e^{- 10 x} + \frac{6}{5} sin (10 x)

y = c_{1} e^{- 10 x} + c_{2} x e^{- 10 x} + \frac{6}{5} sin (10 x)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (2 x), y (0) = 1

y^{^{''}} - y^{^{'}} - 6 y = 61 cos (3 x)

y^{^{'''}} + 6 \cdot y^{^{''}} = x^{2}

y^{^{''}} - 14 y^{^{'}} + 49 y = x^{2}

y^{^{''}} + 16 y^{^{'}} = x + e^{2 x}

What is the solution for y^{''}+20y^'+100y=240cos(10x) ?
The solution for y^{''}+20y^'+100y=240cos(10x) is y=c_{1}e^{-10x}+c_{2}xe^{-10x}+6/5 sin(10x)