What is the solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 is y=(1-8/13)e^{7t}+8/13 e^{-6t}

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y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1

y^{^{''}} - y^{^{'}} - 42 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y = (1 - \frac{8}{13}) e^{7 t} + \frac{8}{13} e^{- 6 t}

Solution steps

y^{''} (t) - y^{'} (t) - 42 y = 0

Solve linear ODE:

y = (1 - \frac{8}{13}) e^{7 t} + \frac{8}{13} e^{- 6 t}

y = (1 - \frac{8}{13}) e^{7 t} + \frac{8}{13} e^{- 6 t}

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 1, y (0) = 1

\frac{1}{8} y^{^{''}} + y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 12

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

x^{^{''}} + a x^{^{'}} = 0

What is the solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-y^'-42y=0,y(0)=1,y^'(0)=-1 is y=(1-8/13)e^{7t}+8/13 e^{-6t}