What is the solution for 12y^{''}-5y^'-3y=e^x ?

The solution for 12y^{''}-5y^'-3y=e^x is y=c_{1}e^{(3x)/4}+c_{2}e^{-x/3}+1/4 e^x

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12y^{''}-5y^'-3y=e^x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = e^{x}

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

Solution steps

12 y^{''} (x) - 5 y^{'} (x) - 3 y = e^{x}

Solve linear ODE:

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

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

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

y^{^{''}} - 2 y^{^{'}} + y = e^{t} + t, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 6 sin (x))

y^{^{''}} + 4 y^{^{'}} = 4 - 4 e^{- 4 x}

y^{^{''}} + y^{^{'}} - 6 y = 4 t e^{2 t}

What is the solution for 12y^{''}-5y^'-3y=e^x ?
The solution for 12y^{''}-5y^'-3y=e^x is y=c_{1}e^{(3x)/4}+c_{2}e^{-x/3}+1/4 e^x