What is the solution for x^{''}-9x=2t^2-cos(t) ?

The solution for x^{''}-9x=2t^2-cos(t) is x=c_{1}e^{3t}+c_{2}e^{-3t}-(2t^2)/9-4/81+1/10 cos(t)

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x^{''}-9x=2t^2-cos(t)

x^{^{''}} - 9 x = 2 t^{2} - cos (t)

x = c_{1} e^{3 t} + c_{2} e^{- 3 t} - \frac{2 t ^{2}}{9} - \frac{4}{81} + \frac{1}{10} cos (t)

Solution steps

x^{''} (t) - 9 x = 2 t^{2} - cos (t)

Solve linear ODE:

x = c_{1} e^{3 t} + c_{2} e^{- 3 t} - \frac{2 t ^{2}}{9} - \frac{4}{81} + \frac{1}{10} cos (t)

x = c_{1} e^{3 t} + c_{2} e^{- 3 t} - \frac{2 t ^{2}}{9} - \frac{4}{81} + \frac{1}{10} cos (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- t}

3 y^{^{'}} + 2 y^{^{''}} = - 10 cos (3.14 \cdot \frac{t}{0.2})

y^{^{''}} + 36 y = sin (4 x)

y^{^{''}} + 3 y = - 48 x^{2} \cdot e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 14 y = 7 + 4 x

What is the solution for x^{''}-9x=2t^2-cos(t) ?
The solution for x^{''}-9x=2t^2-cos(t) is x=c_{1}e^{3t}+c_{2}e^{-3t}-(2t^2)/9-4/81+1/10 cos(t)