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

The solution for 3y^{''}+4y^'-3y=0 is y=c_{1}e^{((-2+sqrt(13))t)/3}+c_{2}e^{-((2+sqrt(13))t)/3}

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

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

y = c_{1} e^{\frac{( - 2 + 13 ) t}{3}} + c_{2} e^{- \frac{( 2 + 13 ) t}{3}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 2 + 13 ) t}{3}} + c_{2} e^{- \frac{( 2 + 13 ) t}{3}}

y = c_{1} e^{\frac{( - 2 + 13 ) t}{3}} + c_{2} e^{- \frac{( 2 + 13 ) t}{3}}

3 y^{^{''}} + 5 y^{^{'}} - y = 0

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

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

2 x^{^{''}} + 10 x = 0

2 y^{^{'''}} - 6 y^{^{''}} = 0

What is the solution for 3y^{''}+4y^'-3y=0 ?
The solution for 3y^{''}+4y^'-3y=0 is y=c_{1}e^{((-2+sqrt(13))t)/3}+c_{2}e^{-((2+sqrt(13))t)/3}