What is the solution for (dp}{dt}=\frac{p-900)/2 ?

The solution for (dp}{dt}=\frac{p-900)/2 is p=e^{(t+c_{1})/2}+900

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(dp}{dt}=\frac{p-900)/2

\frac{d p}{d t} = \frac{p - 900}{2}

p = e^{\frac{t + c _{1}}{2}} + 900

Solution steps

\frac{d p}{d t} = \frac{p - 900}{2}

Solve separable ODE:

p = e^{\frac{t + c _{1}}{2}} + 900

p = e^{\frac{t + c _{1}}{2}} + 900

220 = (0.5) (x) + (15.5) \frac{d x}{d t} + (20)

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

\frac{d x}{d y} = y (4 x^{2} - 1)

\frac{d y}{d x} y = ln ((x^{5} + 7)^{2 π})

\frac{d y}{d x} y = (tan (x))^{l n (x)}

What is the solution for (dp}{dt}=\frac{p-900)/2 ?
The solution for (dp}{dt}=\frac{p-900)/2 is p=e^{(t+c_{1})/2}+900