What is the solution for y^{''}+10y^'+25y=0,y(0)=8,y^'(0)=7 ?

The solution for y^{''}+10y^'+25y=0,y(0)=8,y^'(0)=7 is y=8e^{-5t}+47te^{-5t}

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y^{''}+10y^'+25y=0,y(0)=8,y^'(0)=7

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 8, y^{^{'}} (0) = 7

y = 8 e^{- 5 t} + 47 t e^{- 5 t}

Solution steps

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

Solve linear ODE:

y = 8 e^{- 5 t} + 47 t e^{- 5 t}

y = 8 e^{- 5 t} + 47 t e^{- 5 t}

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = 4, y^{^{'}} (0) = 4

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

(4 D^{4} - 4 D^{3} - 23 D^{2} + 12 D + 36) y = 0

2500 y^{^{''}} - 9 y = 0, y (0) = 6, y^{^{'}} (0) = 7

\frac{d ^{3} x}{d t ^{3}} - x = 0

What is the solution for y^{''}+10y^'+25y=0,y(0)=8,y^'(0)=7 ?
The solution for y^{''}+10y^'+25y=0,y(0)=8,y^'(0)=7 is y=8e^{-5t}+47te^{-5t}