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

The solution for x^{''}-2x^'-3x=0 is x=c_{1}e^{3t}+c_{2}e^{-t}

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x^{''}-2x^'-3x=0

x^{^{''}} - 2 x^{^{'}} - 3 x = 0

x = c_{1} e^{3 t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{3 t} + c_{2} e^{- t}

x = c_{1} e^{3 t} + c_{2} e^{- t}

\frac{d}{d x} (2 x y)

x^{^{'}} = 6 (x^{2} + 1)

\frac{d}{d x} (1 0^{x^{2}})

\int_{0}^{4} \frac{1}{9 + x ^{2}} d x

\frac{\partial}{\partial x} (- 6 x + 12)

What is the solution for x^{''}-2x^'-3x=0 ?
The solution for x^{''}-2x^'-3x=0 is x=c_{1}e^{3t}+c_{2}e^{-t}