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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

77 y^{^{''''}} + 204 y^{^{''}} + 64 y = 0

3 y^{^{''}} - 9 y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} + 10 (\frac{d y}{d x}) + 25 y = 0

x^{^{''}} + 8 x^{^{'}} + 80 x = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 5, y^{^{'}} (- 1) = 4

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