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

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

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y^{''}-4y+4=-5e^{3x}

y^{^{''}} - 4 y + 4 = - 5 e^{3 x}

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 4 x = cos (4 t)

x^{^{''}} - 2 x^{^{'}} + 3 x + 2 = 0

y^{^{''}} - 10 y^{^{'}} + 16 = 0

y^{^{''}} + 2 y^{^{'}} + y = 3 x \cdot e^{- x}

y^{^{'''}} - 12 y^{^{''}} + 44 y^{^{'}} = 48

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