What is the solution for (dθ)/(dt)=40-2/5 (θ-15) ?

The solution for (dθ)/(dt)=40-2/5 (θ-15) is θ(t)=(e^{(-2t-c_{1})/5})/(2)+115

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(dθ)/(dt)=40-2/5 (θ-15)

\frac{d θ}{d t} = 40 - \frac{2}{5} (θ - 15)

θ (t) = \frac{e ^{\frac{- 2 t - c _{1}}{5}}}{2} + 115

Solution steps

\frac{d θ}{d t} = 40 - \frac{2}{5} (θ - 15)

Solve separable ODE:

θ (t) = \frac{e ^{\frac{- 2 t - c _{1}}{5}}}{2} + 115

θ (t) = \frac{e ^{\frac{- 2 t - c _{1}}{5}}}{2} + 115

y^{^{''}} + 4 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

f (x) = (9 x^{2} - 15 x) e^{x}

\int sin (x) cos (x) ln (sin (x)) d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{7}{3}

\int_{0}^{1} 4 - x^{2} d x

What is the solution for (dθ)/(dt)=40-2/5 (θ-15) ?
The solution for (dθ)/(dt)=40-2/5 (θ-15) is θ(t)=(e^{(-2t-c_{1})/5})/(2)+115