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

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

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y^{''}+3y^'-4y=2(5x+1)e^x

y^{^{''}} + 3 y^{^{'}} - 4 y = 2 (5 x + 1) e^{x}

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

Solution steps

y^{''} (x) + 3 y^{'} (x) - 4 y = 2 (5 x + 1) e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 8 cos (3 x) + 12 sin (3 x)

2 y^{^{''}} + 27 y = e^{t} - e^{- t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} + 2 y = 6 e^{\frac{1}{2}}

y^{^{''}} + 16 y = 2 csc^{2} (4 x)

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

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