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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = \frac{4}{1 - 6 x}

n = 1 \sum \infty \frac{4}{3 ^{n - 1}}

x \to 2 lim (lo g_{10 x} (x))

\frac{\partial}{\partial x} (3 x y^{2} - ln (x))

\frac{2 s}{( s + 1 ) ^{2}}

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