What is the solution for (dx)/(dt)=0.3*10-x/(40) ?

The solution for (dx)/(dt)=0.3*10-x/(40) is x=-e^{(-t-c_{1})/(40)}+120

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(dx)/(dt)=0.3*10-x/(40)

\frac{d x}{d t} = 0.3 \cdot 10 - \frac{x}{40}

x = - e^{\frac{- t - c _{1}}{40}} + 120

Solution steps

\frac{d x}{d t} = 0.3 \cdot 10 - \frac{x}{40}

Solve separable ODE:

x = - e^{\frac{- t - c _{1}}{40}} + 120

x = - e^{\frac{- t - c _{1}}{40}} + 120

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\int \frac{1}{3 + sin ^{2} ( x ) - cos ^{2} ( x )} d x

h (x) = \frac{1}{x ^{2} + 4 x - 21}

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\frac{d}{d x} (19)

What is the solution for (dx)/(dt)=0.3*10-x/(40) ?
The solution for (dx)/(dt)=0.3*10-x/(40) is x=-e^{(-t-c_{1})/(40)}+120