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=(3-1/4)e^t+1/4 e^{-3t}

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

Solution steps

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

Solve linear ODE:

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

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

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

25 y^{^{''}} - 30 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{3} \frac{x}{5}

4 y^{^{''}} - 4 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - 1.5

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 3

(d^{4} - 16) 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=(3-1/4)e^t+1/4 e^{-3t}