What is the solution for 2y^{''}+8y+10y=0,y(0)=5,y^'(0)=0 ?

The solution for 2y^{''}+8y+10y=0,y(0)=5,y^'(0)=0 is y=5cos(3t)

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2y^{''}+8y+10y=0,y(0)=5,y^'(0)=0

2 y^{^{''}} + 8 y + 10 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y = 5 cos (3 t)

Solution steps

2 y^{''} (t) + 8 y + 10 y = 0

Solve linear ODE:

y = 5 cos (3 t)

y = 5 cos (3 t)

5 x^{^{''}} + 10 x^{^{'}} + 25 x = 0

u^{^{''}} + 7 u^{^{'}} + 2 H (t) = 0

y^{^{''}} + 6 y^{^{'}} - 16 y = 7 y^{^{'}} - 10 y

y^{^{''}} - \frac{n ^{2}}{4} y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 (\frac{d y}{d x}) + 4 y = 0

What is the solution for 2y^{''}+8y+10y=0,y(0)=5,y^'(0)=0 ?
The solution for 2y^{''}+8y+10y=0,y(0)=5,y^'(0)=0 is y=5cos(3t)