What is the solution for y^{''}+7y^'=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+7y^'=0,y(0)=1,y^'(0)=1 is y=1+1/7-1/7 e^{-7t}

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y^{''}+7y^'=0,y(0)=1,y^'(0)=1

y^{^{''}} + 7 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

y = 1 + \frac{1}{7} - \frac{1}{7} e^{- 7 t}

Solution steps

y^{''} (t) + 7 y^{'} (t) = 0

Solve linear ODE:

y = 1 + \frac{1}{7} - \frac{1}{7} e^{- 7 t}

y = 1 + \frac{1}{7} - \frac{1}{7} e^{- 7 t}

\frac{d y}{d x} = 2 y x + y x^{2}, y (- 3) = 1

h (x) = \frac{1}{2 x}, at x = 4

x \to - 2 lim (2 x^{2})

\int x^{3} 5 + x^{2} d x

x + 7

What is the solution for y^{''}+7y^'=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+7y^'=0,y(0)=1,y^'(0)=1 is y=1+1/7-1/7 e^{-7t}