What is the solution for y^'=0.19y+200 ?

The solution for y^'=0.19y+200 is y=(e^{0.19t+c_{1}})/(0.19)-(200)/(0.19)

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y^'=0.19y+200

y^{^{'}} = 0.19 y + 200

y = \frac{e ^{0.19 t + c_{1}}}{0.19} - \frac{200}{0.19}

Solution steps

y^{'} (t) = 0.19 y + 200

Solve separable ODE:

y = \frac{e ^{0.19 t + c_{1}}}{0.19} - \frac{200}{0.19}

y = \frac{e ^{0.19 t + c_{1}}}{0.19} - \frac{200}{0.19}

\int 5 e^{4} d x

y = x^{\frac{4}{7}} + 6 x

\frac{d}{d x} (x^{4} - 2 x^{3} + 3 x + 1)

\int \frac{2 - 5 e ^{- x}}{e ^{- x}} d x

What is the solution for y^'=0.19y+200 ?
The solution for y^'=0.19y+200 is y=(e^{0.19t+c_{1}})/(0.19)-(200)/(0.19)