What is the solution for y^'=0.0411*y-480 ?

The solution for y^'=0.0411*y-480 is y=(e^{0.0411t+c_{1}})/(0.0411)+(480)/(0.0411)

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y^'=0.0411*y-480

y^{^{'}} = 0.0411 \cdot y - 480

y = \frac{e ^{0.0411 t + c_{1}}}{0.0411} + \frac{480}{0.0411}

Solution steps

y^{'} (t) = 0.0411 \cdot y - 480

Solve separable ODE:

y = \frac{e ^{0.0411 t + c_{1}}}{0.0411} + \frac{480}{0.0411}

y = \frac{e ^{0.0411 t + c_{1}}}{0.0411} + \frac{480}{0.0411}

cos (x) sin (y) d x + sin (x) cos (y) d y = 0

y^{^{'}} = 2 y^{2} + y

\frac{d P}{d t} = 0.7 P - 0.001 P^{2}

(3 x - 3 y^{2} x) d x + (5 y - 5 x^{2} y) d y = 0

x^{2} \frac{d y}{d x} + 3 x = - 3 x y

What is the solution for y^'=0.0411*y-480 ?
The solution for y^'=0.0411*y-480 is y=(e^{0.0411t+c_{1}})/(0.0411)+(480)/(0.0411)