What is the solution for y^{''}+4y+68y=0,y(0)=7 ?

The solution for y^{''}+4y+68y=0,y(0)=7 is y=7cos(6sqrt(2)t)+c_{2}sin(6sqrt(2)t)

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y^{''}+4y+68y=0,y(0)=7

y^{^{''}} + 4 y + 68 y = 0, y (0) = 7

y = 7 cos (62 t) + c_{2} sin (62 t)

Solution steps

y^{''} (t) + 4 y + 68 y = 0

Solve linear ODE:

y = 7 cos (62 t) + c_{2} sin (62 t)

y = 7 cos (62 t) + c_{2} sin (62 t)

3 y^{^{''}} + 48 y = 0, y (0) = 4, y^{^{'}} (0) = 8

y^{^{''''}} + 4 y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} - 2 y = 0

\frac{d ^{2} f}{d x ^{2}} + \frac{df}{d x} = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = - 8

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 10, y^{^{'}} (0) = 27

What is the solution for y^{''}+4y+68y=0,y(0)=7 ?
The solution for y^{''}+4y+68y=0,y(0)=7 is y=7cos(6sqrt(2)t)+c_{2}sin(6sqrt(2)t)