What is the solution for y^{''}-5y^'-6y=0,y(0)=x,y^'(0)=4 ?

The solution for y^{''}-5y^'-6y=0,y(0)=x,y^'(0)=4 is y=c_{1}e^{6x}+c_{2}e^{-x}

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y^{''}-5y^'-6y=0,y(0)=x,y^'(0)=4

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = x, y^{^{'}} (0) = 4

y = c_{1} e^{6 x} + c_{2} e^{- x}

Solution steps

y^{''} (x) - 5 y^{'} (x) - 6 y = 0

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- x}

y = c_{1} e^{6 x} + c_{2} e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - 2

y^{^{''}} + 20 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 24

y^{^{'''}} - 3 y^{^{''}} - 54 y^{^{'}} = 0

77 y^{^{''''}} + 123 y^{^{''}} + 16 y = 0

1.6 x^{^{''}} + 40 x = 0, x (0) = - \frac{1}{3}, x^{^{'}} (0) = \frac{5}{4}

What is the solution for y^{''}-5y^'-6y=0,y(0)=x,y^'(0)=4 ?
The solution for y^{''}-5y^'-6y=0,y(0)=x,y^'(0)=4 is y=c_{1}e^{6x}+c_{2}e^{-x}