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

The solution for y^{''}-2y^'-3y=0,y(0)=3,y^'(0)=2 is y=(3-7/4)e^{3t}+7/4 e^{-t}

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

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

y = (3 - \frac{7}{4}) e^{3 t} + \frac{7}{4} e^{- t}

Solution steps

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

Solve linear ODE:

y = (3 - \frac{7}{4}) e^{3 t} + \frac{7}{4} e^{- t}

y = (3 - \frac{7}{4}) e^{3 t} + \frac{7}{4} e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 15 y = 0

y^{^{''}} + 7 y^{^{'}} - 8 y = 0

y^{^{''}} + 17 y^{^{'}} + 72 y = 0

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - 11 y^{^{'}} + 24 y = 0

What is the solution for y^{''}-2y^'-3y=0,y(0)=3,y^'(0)=2 ?
The solution for y^{''}-2y^'-3y=0,y(0)=3,y^'(0)=2 is y=(3-7/4)e^{3t}+7/4 e^{-t}