What is the solution for 3y^{''}-2y+4y^'=0 ?

The solution for 3y^{''}-2y+4y^'=0 is y=c_{1}e^{((-2+sqrt(10))t)/3}+c_{2}e^{-((2+sqrt(10))t)/3}

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3y^{''}-2y+4y^'=0

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

y = c_{1} e^{\frac{( - 2 + 10 ) t}{3}} + c_{2} e^{- \frac{( 2 + 10 ) t}{3}}

Solution steps

3 y^{''} (t) - 2 y + 4 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 2 + 10 ) t}{3}} + c_{2} e^{- \frac{( 2 + 10 ) t}{3}}

y = c_{1} e^{\frac{( - 2 + 10 ) t}{3}} + c_{2} e^{- \frac{( 2 + 10 ) t}{3}}

y^{^{'}} x + \frac{y ^{^{''}}}{2} x = 0

y^{^{''}} - 7 y^{^{''}} + 19 y^{^{'}} - 13 y = 0

\frac{d ^{2} y}{d x ^{2}} + 2 a \frac{d y}{d x} + (b^{2}) y = 0

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = 3

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 5^{2} y = 0

What is the solution for 3y^{''}-2y+4y^'=0 ?
The solution for 3y^{''}-2y+4y^'=0 is y=c_{1}e^{((-2+sqrt(10))t)/3}+c_{2}e^{-((2+sqrt(10))t)/3}