What is the solution for y^{''}-5y^'+6y=2e^{-x},y(0)=0,y^'(0)=1 ?

The solution for y^{''}-5y^'+6y=2e^{-x},y(0)=0,y^'(0)=1 is y=(5/3-1/6)e^{3x}-5/3 e^{2x}+1/6 e^{-x}

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y^{''}-5y^'+6y=2e^{-x},y(0)=0,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{- x}, y (0) = 0, y^{^{'}} (0) = 1

y = (\frac{5}{3} - \frac{1}{6}) e^{3 x} - \frac{5}{3} e^{2 x} + \frac{1}{6} e^{- x}

Solution steps

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

Solve linear ODE:

y = (\frac{5}{3} - \frac{1}{6}) e^{3 x} - \frac{5}{3} e^{2 x} + \frac{1}{6} e^{- x}

y = (\frac{5}{3} - \frac{1}{6}) e^{3 x} - \frac{5}{3} e^{2 x} + \frac{1}{6} e^{- x}

y^{^{''}} + 5 y = e^{t}

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

y^{^{''}} + y = 1, y^{^{'}} (0) = - 1

y^{^{''}} + y = 1, y^{^{'}} (0) = - 3

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

What is the solution for y^{''}-5y^'+6y=2e^{-x},y(0)=0,y^'(0)=1 ?
The solution for y^{''}-5y^'+6y=2e^{-x},y(0)=0,y^'(0)=1 is y=(5/3-1/6)e^{3x}-5/3 e^{2x}+1/6 e^{-x}