What is the solution for y/(12000)+0.058y^'=0.0015 ?

The solution for y/(12000)+0.058y^'=0.0015 is y=-e^{(-t-c_{1})/(696)}+18

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y/(12000)+0.058y^'=0.0015

\frac{y}{12000} + 0.058 y^{^{'}} = 0.0015

y = - e^{\frac{- t - c _{1}}{696}} + 18

Solution steps

\frac{y}{12000} + 0.058 y^{'} (t) = 0.0015

Solve separable ODE:

y = - e^{\frac{- t - c _{1}}{696}} + 18

y = - e^{\frac{- t - c _{1}}{696}} + 18

x \to \infty lim (- 18 x^{4})

y = - e^{- t}

\int \frac{( x ^{2} + 2 )}{x ^{4} + 4} d x

\frac{d}{d x} (x (x^{2} + 1)^{- \frac{1}{2}})

\int \frac{7 x ^{2} + 5 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

What is the solution for y/(12000)+0.058y^'=0.0015 ?
The solution for y/(12000)+0.058y^'=0.0015 is y=-e^{(-t-c_{1})/(696)}+18