What is the solution for linear x^{''}=-x^'+3x ?

The solution for linear x^{''}=-x^'+3x is x=c_{1}e^{((-1+sqrt(13))t)/2}+c_{2}e^{((-1-sqrt(13))t)/2}

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linear x^{''}=-x^'+3x

linear

x^{^{''}} = - x^{^{'}} + 3 x

x = c_{1} e^{\frac{( - 1 + 13 ) t}{2}} + c_{2} e^{\frac{( - 1 - 13 ) t}{2}}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{\frac{( - 1 + 13 ) t}{2}} + c_{2} e^{\frac{( - 1 - 13 ) t}{2}}

x = c_{1} e^{\frac{( - 1 + 13 ) t}{2}} + c_{2} e^{\frac{( - 1 - 13 ) t}{2}}

y^{^{'''}} - 7 y^{^{''}} + 14 y^{^{'}} - 8 y = 0

y^{^{''}} + 2 y^{^{'}} + y = - y^{^{''}} + 5 y + 2 y^{^{'}} - y

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

y^{^{''}} - 6 y^{^{'}} + 32 y = 0

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = sin (3 x)

What is the solution for linear x^{''}=-x^'+3x ?
The solution for linear x^{''}=-x^'+3x is x=c_{1}e^{((-1+sqrt(13))t)/2}+c_{2}e^{((-1-sqrt(13))t)/2}