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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} - 6 \frac{d x}{d t} - 7 x = 6

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{3} e^{t}

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

y^{^{''}} + 4 y = \frac{t}{2} - \frac{t cos ( 2 t )}{2}

y^{^{''}} - 7 y^{^{'}} - 8 y = 8 x^{2} - 2 x

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