What is the solution for y^{''}+7y^'-8=0 ?

The solution for y^{''}+7y^'-8=0 is y=c_{1}+c_{2}e^{-7t}+(8t)/7

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y^{''}+7y^'-8=0

y^{^{''}} + 7 y^{^{'}} - 8 = 0

y = c_{1} + c_{2} e^{- 7 t} + \frac{8 t}{7}

Solution steps

y^{''} (t) + 7 y^{'} (t) - 8 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 7 t} + \frac{8 t}{7}

y = c_{1} + c_{2} e^{- 7 t} + \frac{8 t}{7}

d (x) = x - a

y^{^{''}} + 2 y^{^{'}} = - 18 x^{2} e^{2 x}

y^{^{''}} + 2 y^{^{'}} - 24 y = - 80 t + 48 t^{2}

y^{^{''}} - 4 y = (x^{3} - 3) sin (2 x)

y^{^{''}} - 2 y^{^{'}} - 63 y = - 181 t + 126 t^{2}

What is the solution for y^{''}+7y^'-8=0 ?
The solution for y^{''}+7y^'-8=0 is y=c_{1}+c_{2}e^{-7t}+(8t)/7