What is the solution for y^{''}+y^'-6y=0,y(0)=3 ?

The solution for y^{''}+y^'-6y=0,y(0)=3 is y=(3-c_{2})e^{2t}+c_{2}e^{-3t}

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y^{''}+y^'-6y=0,y(0)=3

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 3

y = (3 - c_{2}) e^{2 t} + c_{2} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = (3 - c_{2}) e^{2 t} + c_{2} e^{- 3 t}

y = (3 - c_{2}) e^{2 t} + c_{2} e^{- 3 t}

x \frac{d y}{d x} - 2 y = x^{2}

\frac{d}{d y} (- e^{- x y})

\int \frac{25 - x ^{2}}{x} d x

\int \frac{sec ( x ) tan ( x )}{2 sec ( x ) + 3} d x

\int (x + 2) (2 x - 1) d x

What is the solution for y^{''}+y^'-6y=0,y(0)=3 ?
The solution for y^{''}+y^'-6y=0,y(0)=3 is y=(3-c_{2})e^{2t}+c_{2}e^{-3t}