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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d t ^{2}} + 5 y = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + 25 y = 0, y (\frac{π}{10}) = - 3, y (π) = - 8

\frac{d ^{3} y}{d x ^{3}} - 5 y = 0

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