What is the solution for y^{''}+2y^'-3y=6x+5 ?

The solution for y^{''}+2y^'-3y=6x+5 is y=c_{1}e^x+c_{2}e^{-3x}-2x-3

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y^{''}+2y^'-3y=6x+5

y^{^{''}} + 2 y^{^{'}} - 3 y = 6 x + 5

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 6 x + 5

Solve linear ODE:

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

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

5 y^{^{''}} + 4 y^{^{'}} + 4 y = 35 e^{- 4 x}

y^{^{'''}} + y^{^{'}} = x sin (x + 1)

y^{^{''}} + 3 y^{^{'}} + 4 = 0

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

y^{^{''}} + 9 y = 3 sec^{3} (3 x)

What is the solution for y^{''}+2y^'-3y=6x+5 ?
The solution for y^{''}+2y^'-3y=6x+5 is y=c_{1}e^x+c_{2}e^{-3x}-2x-3