What is the solution for 6y^{''}-10y^'-4y=x+2 ?

The solution for 6y^{''}-10y^'-4y=x+2 is y=c_{1}e^{2x}+c_{2}e^{-x/3}-x/4+1/8

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6y^{''}-10y^'-4y=x+2

6 y^{^{''}} - 10 y^{^{'}} - 4 y = x + 2

y = c_{1} e^{2 x} + c_{2} e^{- \frac{x}{3}} - \frac{x}{4} + \frac{1}{8}

Solution steps

6 y^{''} (x) - 10 y^{'} (x) - 4 y = x + 2

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- \frac{x}{3}} - \frac{x}{4} + \frac{1}{8}

y = c_{1} e^{2 x} + c_{2} e^{- \frac{x}{3}} - \frac{x}{4} + \frac{1}{8}

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

y^{^{''}} - y^{^{'}} = x e [x], y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + y = 2 sin (2 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x} + \frac{x}{2}

y^{^{''}} - y^{^{'}} = t^{2}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 6y^{''}-10y^'-4y=x+2 ?
The solution for 6y^{''}-10y^'-4y=x+2 is y=c_{1}e^{2x}+c_{2}e^{-x/3}-x/4+1/8