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

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

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

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

x = (2 + \frac{1}{4}) e^{t} - \frac{1}{4} e^{- 3 t}

Solution steps

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

Solve linear ODE:

x = (2 + \frac{1}{4}) e^{t} - \frac{1}{4} e^{- 3 t}

x = (2 + \frac{1}{4}) e^{t} - \frac{1}{4} e^{- 3 t}

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - 6

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} - y^{^{'}} - 72 y^{^{'}} = 0

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

\frac{y ^{^{''}}}{8} + 30 y^{^{'}} + 500 y = 0

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