What is the solution for y^{''}-4y^'-5y=-2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y^'-5y=-2,y(0)=0,y^'(0)=0 is y=(1/3-2/5)e^{5t}-1/3 e^{-t}+2/5

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y^{''}-4y^'-5y=-2,y(0)=0,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} - 5 y = - 2, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = - 2

Solve linear ODE:

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

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

What is the solution for y^{''}-4y^'-5y=-2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'-5y=-2,y(0)=0,y^'(0)=0 is y=(1/3-2/5)e^{5t}-1/3 e^{-t}+2/5