What is the solution for y^{''}+1.9*y^'=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+1.9*y^'=0,y(0)=0,y^'(0)=1 is y= 10/19-10/19 e^{-(19t)/(10)}

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y^{''}+1.9*y^'=0,y(0)=0,y^'(0)=1

y^{^{''}} + 1.9 \cdot y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{10}{19} - \frac{10}{19} e^{- \frac{19 t}{10}}

Solution steps

y^{''} (t) + 1.9 \cdot y^{'} (t) = 0

Solve linear ODE:

y = \frac{10}{19} - \frac{10}{19} e^{- \frac{19 t}{10}}

y = \frac{10}{19} - \frac{10}{19} e^{- \frac{19 t}{10}}

\int \frac{x}{x ^{2} + x - 1} d x

\frac{d}{d x} (sec (8 x^{3} + 1))

f (x) = 18 x, at x = 9

n = 2 \sum \infty \frac{7}{4 ^{n}}

x \to \infty lim (e^{x} x)

What is the solution for y^{''}+1.9*y^'=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+1.9*y^'=0,y(0)=0,y^'(0)=1 is y= 10/19-10/19 e^{-(19t)/(10)}