What is the solution for 3y^{''}+12y^'-2y=0 ?

The solution for 3y^{''}+12y^'-2y=0 is y=c_{1}e^{((-6+sqrt(42))t)/3}+c_{2}e^{-((6+sqrt(42))t)/3}

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3y^{''}+12y^'-2y=0

3 y^{^{''}} + 12 y^{^{'}} - 2 y = 0

y = c_{1} e^{\frac{( - 6 + 42 ) t}{3}} + c_{2} e^{- \frac{( 6 + 42 ) t}{3}}

Solution steps

3 y^{''} (t) + 12 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 6 + 42 ) t}{3}} + c_{2} e^{- \frac{( 6 + 42 ) t}{3}}

y = c_{1} e^{\frac{( - 6 + 42 ) t}{3}} + c_{2} e^{- \frac{( 6 + 42 ) t}{3}}

y^{^{''}} + 200 y^{^{'}} + 1000 y = 0

y^{^{''}} - 4^{2} y = 0

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

y^{^{''}} + 12 y^{^{'}} + 32 y = 0

\frac{d ^{2} x}{d t ^{2}} + \frac{53}{49} \frac{d x}{d t} + 49 x = 0

What is the solution for 3y^{''}+12y^'-2y=0 ?
The solution for 3y^{''}+12y^'-2y=0 is y=c_{1}e^{((-6+sqrt(42))t)/3}+c_{2}e^{-((6+sqrt(42))t)/3}