What is the solution for y^'=-9.8-(0.0025)/4 y ?

The solution for y^'=-9.8-(0.0025)/4 y is y=(e^{\frac{-t-c_{1})/(1600)}}{0.000625}-(9.8)/(0.000625)

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y^'=-9.8-(0.0025)/4 y

y^{^{'}} = - 9.8 - \frac{0.0025}{4} y

y = \frac{e ^{\frac{- t - c _{1}}{1600}}}{0.000625} - \frac{9.8}{0.000625}

Solution steps

y^{'} (t) = - 9.8 - \frac{0.0025}{4} y

Solve separable ODE:

y = \frac{e ^{\frac{- t - c _{1}}{1600}}}{0.000625} - \frac{9.8}{0.000625}

y = \frac{e ^{\frac{- t - c _{1}}{1600}}}{0.000625} - \frac{9.8}{0.000625}

y^{^{'}} = y tan (x)

\frac{d x}{d t} = t e^{x}, x = 1

\frac{d y}{d x} = 2 x^{3} y^{4}

\frac{( d y )}{( d x ) - 3 y} = x^{(6)}

\frac{d x}{sin ( y )} + \frac{d y}{cos ( x )} = 0

What is the solution for y^'=-9.8-(0.0025)/4 y ?
The solution for y^'=-9.8-(0.0025)/4 y is y=(e^{\frac{-t-c_{1})/(1600)}}{0.000625}-(9.8)/(0.000625)