What is the solution for y^{''}+10y^'+25y=0,y(0)=10 ?

The solution for y^{''}+10y^'+25y=0,y(0)=10 is y=10e^{-5t}+c_{2}te^{-5t}

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

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

y = 10 e^{- 5 t} + c_{2} t e^{- 5 t}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = 10 e^{- 5 t} + c_{2} t e^{- 5 t}

y = 10 e^{- 5 t} + c_{2} t e^{- 5 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = - \frac{28}{3}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 3.65

y^{^{''}} + 2 y^{^{'}} = 0, y (0) = - 3, y^{^{'}} (0) = 12

y^{^{'''}} - 4 y^{^{'}} - 5 y^{^{'}} = 0

0.5 x^{^{''}} + 0.5 x^{^{'}} + 6 x = 0, x (0) = 3, x^{^{'}} (0) = 0

What is the solution for y^{''}+10y^'+25y=0,y(0)=10 ?
The solution for y^{''}+10y^'+25y=0,y(0)=10 is y=10e^{-5t}+c_{2}te^{-5t}