What is the solution for y^{''}-y^'-2y=e^{3x},y(0)=1,y^'(0)=2 ?

The solution for y^{''}-y^'-2y=e^{3x},y(0)=1,y^'(0)=2 is y=(-1/12+3/4)e^{2x}+1/12 e^{-x}+1/4 e^{3x}

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y^{''}-y^'-2y=e^{3x},y(0)=1,y^'(0)=2

y^{^{''}} - y^{^{'}} - 2 y = e^{3 x}, y (0) = 1, y^{^{'}} (0) = 2

y = (- \frac{1}{12} + \frac{3}{4}) e^{2 x} + \frac{1}{12} e^{- x} + \frac{1}{4} e^{3 x}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = e^{3 x}

Solve linear ODE:

y = (- \frac{1}{12} + \frac{3}{4}) e^{2 x} + \frac{1}{12} e^{- x} + \frac{1}{4} e^{3 x}

y = (- \frac{1}{12} + \frac{3}{4}) e^{2 x} + \frac{1}{12} e^{- x} + \frac{1}{4} e^{3 x}

y^{^{''}} - 2 y^{^{'}} + y = 6

y^{^{''}} - 2 y^{^{'}} + y = 4

y^{^{''}} + 4 y^{^{'}} - 5 y = 35 e^{- 4 t}

y^{^{''}} + 3 y = - 4 x^{2} e^{3 x}

y^{^{''}} + 9 y = 27 x^{3}

What is the solution for y^{''}-y^'-2y=e^{3x},y(0)=1,y^'(0)=2 ?
The solution for y^{''}-y^'-2y=e^{3x},y(0)=1,y^'(0)=2 is y=(-1/12+3/4)e^{2x}+1/12 e^{-x}+1/4 e^{3x}