What is the solution for y^{''}+14y^'+50y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+14y^'+50y=0,y(0)=1,y^'(0)=0 is y=e^{-7t}(cos(t)+7sin(t))

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y^{''}+14y^'+50y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 14 y^{^{'}} + 50 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = e^{- 7 t} (cos (t) + 7 sin (t))

Solution steps

y^{''} (t) + 14 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{- 7 t} (cos (t) + 7 sin (t))

y = e^{- 7 t} (cos (t) + 7 sin (t))

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 5, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 13 \frac{d y}{d x} + 12 y = 0

\frac{3}{4} x^{^{''}} + 72 x = 0, x (0) = \frac{1}{3}, x^{^{'}} (0) = 0

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

y^{^{''}} + 81 y = 0, y (0) = 1, y^{^{'}} (0) = 7

What is the solution for y^{''}+14y^'+50y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+14y^'+50y=0,y(0)=1,y^'(0)=0 is y=e^{-7t}(cos(t)+7sin(t))