What is the solution for θ^{''}=3θ ?

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

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θ^{''}=3θ

θ^{^{''}} = 3 θ

θ (t) = c_{1} e^{3 t} + c_{2} e^{- 3 t}

Solution steps

θ^{''} (t) = 3 θ (t)

Solve linear ODE:

θ (t) = c_{1} e^{3 t} + c_{2} e^{- 3 t}

θ (t) = c_{1} e^{3 t} + c_{2} e^{- 3 t}

\int (2 u + 1)^{2} d u

f (x) = \frac{x}{1 + x ^{2}}, at x = 3

5 x^{2} + 3 x^{4}

\frac{( 2 )}{( s - 1 ) ^{2}}

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