What is the solution for y^{''}+6y+25y=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+6y+25y=0,y(0)=1,y^'(0)=-1 is y=cos(sqrt(31)t)-1/(sqrt(31))sin(sqrt(31)t)

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y^{''}+6y+25y=0,y(0)=1,y^'(0)=-1

y^{^{''}} + 6 y + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y = cos (31 t) - \frac{1}{31} sin (31 t)

Solution steps

y^{''} (t) + 6 y + 25 y = 0

Solve linear ODE:

y = cos (31 t) - \frac{1}{31} sin (31 t)

y = cos (31 t) - \frac{1}{31} sin (31 t)

8 y^{^{''}} + 2 y^{^{'}} = 0

15 x^{^{''}} + 44 x = 0

x^{^{''}} - 7 x^{^{'}} + 10 x = 0

\frac{3}{4} x^{^{''}} + 9 x = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 6, y^{^{'}} (0) = 1

What is the solution for y^{''}+6y+25y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+6y+25y=0,y(0)=1,y^'(0)=-1 is y=cos(sqrt(31)t)-1/(sqrt(31))sin(sqrt(31)t)