What is the solution for y^{''}+2/5 y^'+1/25 y=0,y(0)=0,y^'(0)=4 ?

The solution for y^{''}+2/5 y^'+1/25 y=0,y(0)=0,y^'(0)=4 is y=4e^{-t/5}t

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y^{''}+2/5 y^'+1/25 y=0,y(0)=0,y^'(0)=4

y^{^{''}} + \frac{2}{5} y^{^{'}} + \frac{1}{25} y = 0, y (0) = 0, y^{^{'}} (0) = 4

y = 4 e^{- \frac{t}{5}} t

Solution steps

y^{''} (t) + \frac{2}{5} y^{'} (t) + \frac{1}{25} y = 0

Solve linear ODE:

y = 4 e^{- \frac{t}{5}} t

y = 4 e^{- \frac{t}{5}} t

y^{^{''''}} - y^{^{''}} = 2 (4 y^{^{'''}})

100 y^{^{''}} + 240 y^{^{'}} + (196 π^{2} + 144) y = 0

y^{^{''}} - 4 y = 0, y^{^{'}} (- 1) = 1, y (2) = 2

y^{^{'''}} + y^{^{''}} + y^{^{'}} - 39 y = 0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 3, y^{^{'}} (0) = 16

What is the solution for y^{''}+2/5 y^'+1/25 y=0,y(0)=0,y^'(0)=4 ?
The solution for y^{''}+2/5 y^'+1/25 y=0,y(0)=0,y^'(0)=4 is y=4e^{-t/5}t