What is the solution for x^{''}+bx^'-c=0 ?

The solution for x^{''}+bx^'-c=0 is x=c_{1}+c_{2}e^{-bt}+(ct)/b

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x^{''}+bx^'-c=0

x^{^{''}} + b x^{^{'}} - c = 0

x = c_{1} + c_{2} e^{- b t} + \frac{c t}{b}

Solution steps

x^{''} (t) + b x^{'} (t) - c = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- b t} + \frac{c t}{b}

x = c_{1} + c_{2} e^{- b t} + \frac{c t}{b}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = x^{2}

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

y^{^{''}} + b y^{^{'}} + cy = e^{x} sin (x)

10 y^{^{''}} + 50 y^{^{'}} + 57.6 y = cos (x)

y^{^{''}} + y = 5 sin (4 x)

What is the solution for x^{''}+bx^'-c=0 ?
The solution for x^{''}+bx^'-c=0 is x=c_{1}+c_{2}e^{-bt}+(ct)/b