What is the solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 is y=1+1/3-1/3 e^{-3t}

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y^{''}+3y^'=0,y(0)=1,y^'(0)=1

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

y = 1 + \frac{1}{3} - \frac{1}{3} e^{- 3 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) = 0

Solve linear ODE:

y = 1 + \frac{1}{3} - \frac{1}{3} e^{- 3 t}

y = 1 + \frac{1}{3} - \frac{1}{3} e^{- 3 t}

d er i v a t i v e 3 x^{\frac{3}{2}}

s l o p e (2, 3), (7, - 6)

\int (3 x - \frac{1}{x})^{2} d x

t an g e n t f (x) = \frac{1}{x + 1}, (0, 1)

a re a y = 6 e^{x}, y = 6 e^{- x}, x = 1

What is the solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+3y^'=0,y(0)=1,y^'(0)=1 is y=1+1/3-1/3 e^{-3t}