What is the solution for -18y^{''}-14y^'=0,y(3)=-3,y^'(3)=-3 ?

The solution for -18y^{''}-14y^'=0,y(3)=-3,y^'(3)=-3 is y=(27e^{7/3})/7 e^{-(7t)/9}-48/7

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-18y^{''}-14y^'=0,y(3)=-3,y^'(3)=-3

- 18 y^{^{''}} - 14 y^{^{'}} = 0, y (3) = - 3, y^{^{'}} (3) = - 3

y = \frac{27 e ^{\frac{7}{3}}}{7} e^{- \frac{7 t}{9}} - \frac{48}{7}

Solution steps

- 18 y^{''} (t) - 14 y^{'} (t) = 0

Solve linear ODE:

y = \frac{27 e ^{\frac{7}{3}}}{7} e^{- \frac{7 t}{9}} - \frac{48}{7}

y = \frac{27 e ^{\frac{7}{3}}}{7} e^{- \frac{7 t}{9}} - \frac{48}{7}

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What is the solution for -18y^{''}-14y^'=0,y(3)=-3,y^'(3)=-3 ?
The solution for -18y^{''}-14y^'=0,y(3)=-3,y^'(3)=-3 is y=(27e^{7/3})/7 e^{-(7t)/9}-48/7