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

The solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=1 is y=(1-1/5)e^{2t}+1/5 e^{-3t}

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

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

y = (1 - \frac{1}{5}) e^{2 t} + \frac{1}{5} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{1}{5}) e^{2 t} + \frac{1}{5} e^{- 3 t}

y = (1 - \frac{1}{5}) e^{2 t} + \frac{1}{5} e^{- 3 t}

1 + 2 x^{2}

ln (3 + x)

\frac{d y}{d x} + y = e^{4 x}

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

x \to 3 lim (3 x - 7)

What is the solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=1 is y=(1-1/5)e^{2t}+1/5 e^{-3t}