What is the solution for 4y^{''}-y=12x+6x ?

The solution for 4y^{''}-y=12x+6x is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-18x

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4y^{''}-y=12x+6x

4 y^{^{''}} - y = 12 x + 6 x

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - 18 x

Solution steps

4 y^{''} (x) - y = 12 x + 6 x

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - 18 x

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - 18 x

y^{^{''}} + 8 y^{^{'}} + 16 y = 3 e^{- 4 t}

1 x^{^{''}} + 2 x = 8 sin (2 t)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = e^{- 2 x}

y^{^{''}} - 8 y^{^{'}} + 16 = 10 e^{4 x} sec (x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 4 t e^{4 t}

What is the solution for 4y^{''}-y=12x+6x ?
The solution for 4y^{''}-y=12x+6x is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-18x