What is the solution for y^{''}+6y^'=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+6y^'=0,y(0)=1,y^'(0)=1 is y=1+1/6-1/6 e^{-6t}

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y^{''}+6y^'=0,y(0)=1,y^'(0)=1

y^{^{''}} + 6 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

y = 1 + \frac{1}{6} - \frac{1}{6} e^{- 6 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) = 0

Solve linear ODE:

y = 1 + \frac{1}{6} - \frac{1}{6} e^{- 6 t}

y = 1 + \frac{1}{6} - \frac{1}{6} e^{- 6 t}

\int x \cdot e^{- 3 x} d x

\int \frac{x ^{2} + 6 x + 5}{x ^{2} + 2 x + 1} d x

\frac{d}{d x} (arccos (\frac{( x + 1 )}{2}))

\int_{- 7}^{0} \frac{1}{9 - x} d x

\frac{d}{d x} (\frac{x ^{5}}{x ^{3}})

What is the solution for y^{''}+6y^'=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+6y^'=0,y(0)=1,y^'(0)=1 is y=1+1/6-1/6 e^{-6t}