What is the solution for x^'=2+x+1/4 x-1/2 ?

The solution for x^'=2+x+1/4 x-1/2 is x=(e^{(5t+c_{1})/4})/(5)-6/5

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x^'=2+x+1/4 x-1/2

x^{^{'}} = 2 + x + \frac{1}{4} x - \frac{1}{2}

x = \frac{e ^{\frac{5 t + c _{1}}{4}}}{5} - \frac{6}{5}

Solution steps

x^{'} (t) = 2 + x + \frac{1}{4} x - \frac{1}{2}

Solve separable ODE:

x = \frac{e ^{\frac{5 t + c _{1}}{4}}}{5} - \frac{6}{5}

x = \frac{e ^{\frac{5 t + c _{1}}{4}}}{5} - \frac{6}{5}

e^{y} \cdot y^{^{'}} = \frac{3 x ^{2}}{1 + y}

\frac{d x}{d t} = - \frac{x}{100} + 125

x^{^{'}} = - x^{2}, x (0) = 5

\frac{d y}{d t} = y^{2} e^{- t}, y (0) = \frac{1}{2}, t \geq 0

f (x) = \frac{1}{1 + x ^{2}} f^{^{'}} (x)

What is the solution for x^'=2+x+1/4 x-1/2 ?
The solution for x^'=2+x+1/4 x-1/2 is x=(e^{(5t+c_{1})/4})/(5)-6/5