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

The solution for y^{''}+y^'-6y=0,y(0)=1 is y=(1-c_{2})e^{2t}+c_{2}e^{-3t}

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

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

y = (1 - c_{2}) e^{2 t} + c_{2} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = (1 - c_{2}) e^{2 t} + c_{2} e^{- 3 t}

y = (1 - c_{2}) e^{2 t} + c_{2} e^{- 3 t}

\frac{\partial}{\partial x} (4 x (x^{2} + y^{2}) - 4 x)

t \to 1 lim (sin (3 π t))

y = x^{2} - 7, (- 5, 18)

y^{^{'}} = \frac{2 x + sec ^{2} ( x )}{2 y}

x^{2} sin (3 x)

What is the solution for y^{''}+y^'-6y=0,y(0)=1 ?
The solution for y^{''}+y^'-6y=0,y(0)=1 is y=(1-c_{2})e^{2t}+c_{2}e^{-3t}