What is the solution for y^{''}+6y+8=0,y(0)=0,y^'(0)=12 ?

The solution for y^{''}+6y+8=0,y(0)=0,y^'(0)=12 is y= 4/3 cos(sqrt(6)t)+2sqrt(6)sin(sqrt(6)t)-4/3

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y^{''}+6y+8=0,y(0)=0,y^'(0)=12

y^{^{''}} + 6 y + 8 = 0, y (0) = 0, y^{^{'}} (0) = 12

y = \frac{4}{3} cos (6 t) + 26 sin (6 t) - \frac{4}{3}

Solution steps

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

Solve linear ODE:

y = \frac{4}{3} cos (6 t) + 26 sin (6 t) - \frac{4}{3}

y = \frac{4}{3} cos (6 t) + 26 sin (6 t) - \frac{4}{3}

f (x) = 4 x^{6} - 5 x^{- 5} + 3 x - \frac{6}{x ^{4}} - 5 e^{x}

\frac{d}{d x} (cos (3 x + 2) - (2 x + 1)^{- 3})

\int \frac{sin ( y )}{x} d x

f (x) = \frac{r}{r ^{2} + 5}

What is the solution for y^{''}+6y+8=0,y(0)=0,y^'(0)=12 ?
The solution for y^{''}+6y+8=0,y(0)=0,y^'(0)=12 is y= 4/3 cos(sqrt(6)t)+2sqrt(6)sin(sqrt(6)t)-4/3