What is the solution for y^{''}-7y^'+6y=2,y^'(0)=7 ?

The solution for y^{''}-7y^'+6y=2,y^'(0)=7 is y=(e^{6t}(7-c_{2}))/6+c_{2}e^t+1/3

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y^{''}-7y^'+6y=2,y^'(0)=7

y^{^{''}} - 7 y^{^{'}} + 6 y = 2, y^{^{'}} (0) = 7

y = \frac{e ^{6 t} ( 7 - c _{2} )}{6} + c_{2} e^{t} + \frac{1}{3}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 6 y = 2

Solve linear ODE:

y = \frac{e ^{6 t} ( 7 - c _{2} )}{6} + c_{2} e^{t} + \frac{1}{3}

y = \frac{e ^{6 t} ( 7 - c _{2} )}{6} + c_{2} e^{t} + \frac{1}{3}

y^{^{'''}} - y^{^{'}} = 1 - x^{2}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{(3 t)}

y^{^{''}} + 4 y = 3 x

y^{^{''}} + 2 a y^{^{'}} + b^{2} y = sin (c t)

x^{^{''}} + 10 x^{^{'}} + 16 x = 2581 cos (5 t)

What is the solution for y^{''}-7y^'+6y=2,y^'(0)=7 ?
The solution for y^{''}-7y^'+6y=2,y^'(0)=7 is y=(e^{6t}(7-c_{2}))/6+c_{2}e^t+1/3