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

\frac{d S}{d t} = 0.02 - \frac{S}{490}

x d x + y d y = x y (x d y + y d x)

y^{^{'}} \cdot ln (y) - y = 0

\frac{x ^{2}}{y} - 4 x e^{y^{2}} \frac{d y}{d x} = 0

y^{^{'}} = \frac{( x ^{2} - 5 )}{y ^{2}}, y (0) = 3

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