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

The solution for y^{''}-5y^'-6y=0,y(0)=a,y^'(0)=4 is y=e^{6t}(a+(4-6a)/7)-(e^{-t}(4-6a))/7

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

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

y = e^{6 t} (a + \frac{4 - 6 a}{7}) - \frac{e ^{- t} ( 4 - 6 a )}{7}

Solution steps

y^{''} (t) - 5 y^{'} (t) - 6 y = 0

Solve linear ODE:

y = e^{6 t} (a + \frac{4 - 6 a}{7}) - \frac{e ^{- t} ( 4 - 6 a )}{7}

y = e^{6 t} (a + \frac{4 - 6 a}{7}) - \frac{e ^{- t} ( 4 - 6 a )}{7}

y^{^{''''}} - 44 y^{^{'''}} + 484 y^{^{''}} = 0

81 y^{^{''}} - 54 y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 1

324 y^{^{''}} + 252 y^{^{'}} + 49 y = 0, y (0) = 9

y^{^{'''}} + 4 y^{^{''}} - 128 y = 0, y (0) = e^{4 x}

y^{^{''}} + 64 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{1}{3}

What is the solution for y^{''}-5y^'-6y=0,y(0)=a,y^'(0)=4 ?
The solution for y^{''}-5y^'-6y=0,y(0)=a,y^'(0)=4 is y=e^{6t}(a+(4-6a)/7)-(e^{-t}(4-6a))/7