What is the solution for (dx)/(dt)=1-x/(40) ?

The solution for (dx)/(dt)=1-x/(40) is x=-e^{(-t-c_{1})/(40)}+40

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(dx)/(dt)=1-x/(40)

\frac{d x}{d t} = 1 - \frac{x}{40}

x = - e^{\frac{- t - c _{1}}{40}} + 40

Solution steps

\frac{d x}{d t} = 1 - \frac{x}{40}

Solve separable ODE:

x = - e^{\frac{- t - c _{1}}{40}} + 40

x = - e^{\frac{- t - c _{1}}{40}} + 40

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (5)

y^{^{'}} = e^{2 y} sin (x), y (0) = 0

\frac{d y}{y} = - \frac{2 x}{x ^{2} - 1} d x

2 x d x + sin (y) d y = 0

\frac{d y}{d x} = \frac{x + 1}{y ^{2}}

What is the solution for (dx)/(dt)=1-x/(40) ?
The solution for (dx)/(dt)=1-x/(40) is x=-e^{(-t-c_{1})/(40)}+40