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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 12 y^{^{''}} + 256 y = e^{8 x}

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

\frac{1}{2} x^{^{''}} + \frac{1}{2} x^{^{'}} + 6 x = 10 cos (3 t)

y^{^{''}} - y^{^{'}} = 4 cos (2 x)

8 y^{^{''}} + 4 y^{^{'}} + 5 y = 1, y (0) = 1

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