What is the solution for 6y^{''}+8y^'+y=0,y(-2)=-3,y^'(-2)=-4 ?

The solution for 6y^{''}+8y^'+y=0,y(-2)=-3,y^'(-2)=-4 is

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6y^{''}+8y^'+y=0,y(-2)=-3,y^'(-2)=-4

6 y^{^{''}} + 8 y^{^{'}} + y = 0, y (- 2) = - 3, y^{^{'}} (- 2) = - 4

y = (- 3 e^{\frac{- 4 + 10}{3}} - \frac{e ^{\frac{10 - 4}{3}} ( 36 - 3 10 )}{2 10}) e^{\frac{( - 4 + 10 ) t}{6}} + \frac{( 36 - 3 10 ) e ^{\frac{- ( 4 + 10 ) t - 2 ( 4 + 10 )}{6}}}{2 10}

Solution steps

6 y^{''} (t) + 8 y^{'} (t) + y = 0

Solve linear ODE:

y = (- 3 e^{\frac{- 4 + 10}{3}} - \frac{e ^{\frac{10 - 4}{3}} ( 36 - 3 10 )}{2 10}) e^{\frac{( - 4 + 10 ) t}{6}} + \frac{( 36 - 3 10 ) e ^{\frac{- ( 4 + 10 ) t - 2 ( 4 + 10 )}{6}}}{2 10}

y = (- 3 e^{\frac{- 4 + 10}{3}} - \frac{e ^{\frac{10 - 4}{3}} ( 36 - 3 10 )}{2 10}) e^{\frac{( - 4 + 10 ) t}{6}} + \frac{( 36 - 3 10 ) e ^{\frac{- ( 4 + 10 ) t - 2 ( 4 + 10 )}{6}}}{2 10}

\int \frac{7 x + 1}{6 x ^{2} + x - 1} d x

x \to 4 lim (- \frac{15}{x - 4})

\frac{\partial}{\partial x} (sin (π x))

\frac{d}{d x} (4 x^{5})

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{5 x}

What is the solution for 6y^{''}+8y^'+y=0,y(-2)=-3,y^'(-2)=-4 ?
The solution for 6y^{''}+8y^'+y=0,y(-2)=-3,y^'(-2)=-4 is