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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int 4 sin (5 x) + 5 cos (3 x) d x

\frac{d y}{d t} = 2 y - t

\frac{d}{d x} ((1 + x^{2})^{5})

\frac{d}{d x} (sin (2 x) \cdot 2)

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