What is the solution for y^{''}-4y^'+3y=-e^{-s},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y^'+3y=-e^{-s},y(0)=0,y^'(0)=0 is y(s)=(-1/4+1/8)e^{3s}+1/4 e^s-1/8 e^{-s}

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y^{''}-4y^'+3y=-e^{-s},y(0)=0,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 3 y = - e^{- s}, y (0) = 0, y^{^{'}} (0) = 0

y (s) = (- \frac{1}{4} + \frac{1}{8}) e^{3 s} + \frac{1}{4} e^{s} - \frac{1}{8} e^{- s}

Solution steps

y^{''} (s) - 4 y^{'} (s) + 3 y (s) = - e^{- s}

Solve linear ODE:

y (s) = (- \frac{1}{4} + \frac{1}{8}) e^{3 s} + \frac{1}{4} e^{s} - \frac{1}{8} e^{- s}

y (s) = (- \frac{1}{4} + \frac{1}{8}) e^{3 s} + \frac{1}{4} e^{s} - \frac{1}{8} e^{- s}

arcsec (2 x + 2)

\int 8 sin^{4} (x) cos^{2} (x) d x

\int e^{(2 x + 1)} d x

x y^{^{'}} - y - 3 = 0

x \to 0 lim (\frac{2 tan ( x )}{x})

What is the solution for y^{''}-4y^'+3y=-e^{-s},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'+3y=-e^{-s},y(0)=0,y^'(0)=0 is y(s)=(-1/4+1/8)e^{3s}+1/4 e^s-1/8 e^{-s}