What is the solution for y^{''}+6y-2y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+6y-2y=0,y(0)=1,y^'(0)=2 is y=cos(2t)+sin(2t)

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y^{''}+6y-2y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 6 y - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = cos (2 t) + sin (2 t)

Solution steps

y^{''} (t) + 6 y - 2 y = 0

Solve linear ODE:

y = cos (2 t) + sin (2 t)

y = cos (2 t) + sin (2 t)

y^{^{''}} - 11 y + 18 y = 0

y^{^{''}} + v y^{^{'}} = 0

2 x^{^{''}} + 11 x^{^{'}} + 5 x = 0

3 u^{^{''}} + 38 u^{^{'}} + 144 H (t) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - 1, y^{^{'}} (0) = - \frac{7}{2}

What is the solution for y^{''}+6y-2y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+6y-2y=0,y(0)=1,y^'(0)=2 is y=cos(2t)+sin(2t)