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

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

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

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

y = - e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = - e^{- 3 t}

y = - e^{- 3 t}

y^{^{''}} - 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

25 y^{^{''}} - y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 13 y = 0

8 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

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

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