What is the solution for y^{''}-2y^'-3y=0,y(0)=1,y^'(0)=0 ?

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

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y^{''}-2y^'-3y=0,y(0)=1,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 0

Solve linear ODE:

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

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

(D^{2} - 3 D - 10) y = 0

y^{^{''''}} + k^{2} \cdot y^{^{''}} = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

y^{^{''}} - a \cdot y^{^{'}} - b \cdot y = 0

9 y^{^{''}} - 12 y^{^{'}} + 29 y = 0, y (0) = 1, y^{^{'}} (1) = 3

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