What is the solution for y^'+((10^3))/4 y+2*10^3=0 ?

The solution for y^'+((10^3))/4 y+2*10^3=0 is y=e^{-250t+c_{1}}-8

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y^'+((10^3))/4 y+2*10^3=0

y^{^{'}} + \frac{( 1 0 ^{3} )}{4} y + 2 \cdot 1 0^{3} = 0

y = e^{- 250 t + c_{1}} - 8

Solution steps

y^{'} (t) + \frac{( 1 0 ^{3} )}{4} y + 2 \cdot 1 0^{3} = 0

Solve separable ODE:

y = e^{- 250 t + c_{1}} - 8

y = e^{- 250 t + c_{1}} - 8

\frac{d y}{d x} = 8 x 9 - y^{2}

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

y^{^{'}} = \frac{x ^{2} - 2}{y}

y^{^{'}} + \frac{y ^{^{'}}}{x} + (1 - \frac{1}{4 x ^{2}}) y = 0

y y^{^{'}} - sin (x) = 0, y (π) = 1

What is the solution for y^'+((10^3))/4 y+2*10^3=0 ?
The solution for y^'+((10^3))/4 y+2*10^3=0 is y=e^{-250t+c_{1}}-8