What is the solution for 3.03y^'+0.755y=6.05 ?

The solution for 3.03y^'+0.755y=6.05 is y=-(e^{-0.24917…t-c_{1}})/(0.755)+(6.05)/(0.755)

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3.03y^'+0.755y=6.05

3.03 y^{^{'}} + 0.755 y = 6.05

y = - \frac{e ^{- 0.24917 \dots t - c_{1}}}{0.755} + \frac{6.05}{0.755}

Solution steps

3.03 y^{'} (t) + 0.755 y = 6.05

Solve separable ODE:

y = - \frac{e ^{- 0.24917 \dots t - c_{1}}}{0.755} + \frac{6.05}{0.755}

y = - \frac{e ^{- 0.24917 \dots t - c_{1}}}{0.755} + \frac{6.05}{0.755}

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

\frac{d x}{3 sin ( y )} = \frac{4 d y}{5 sin ( x )}

x e^{- y} d y + \frac{x ^{2} + 1}{y} d x = 0

y^{^{'}} y^{1} = - 2 e^{x} y^{2}

\frac{d y}{d x} = y \cdot sin^{3} (x)

What is the solution for 3.03y^'+0.755y=6.05 ?
The solution for 3.03y^'+0.755y=6.05 is y=-(e^{-0.24917…t-c_{1}})/(0.755)+(6.05)/(0.755)