What is the solution for 220=(0.5)(x)+(15.5)(dx)/(dt)+(20) ?

The solution for 220=(0.5)(x)+(15.5)(dx)/(dt)+(20) is x=2e^{(-t-c_{1})/(31)}+400

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220=(0.5)(x)+(15.5)(dx)/(dt)+(20)

220 = (0.5) (x) + (15.5) \frac{d x}{d t} + (20)

x = 2 e^{\frac{- t - c _{1}}{31}} + 400

Solution steps

220 = (0.5) (x) + (15.5) \frac{d x}{d t} + (20)

Solve separable ODE:

x = 2 e^{\frac{- t - c _{1}}{31}} + 400

x = 2 e^{\frac{- t - c _{1}}{31}} + 400

x^{3} y^{^{'}} = y^{2} (x - 4)

\frac{d x}{d y} = y (4 x^{2} - 1)

\frac{d y}{d x} y = ln ((x^{5} + 7)^{2 π})

\frac{d y}{d x} y = (tan (x))^{l n (x)}

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

What is the solution for 220=(0.5)(x)+(15.5)(dx)/(dt)+(20) ?
The solution for 220=(0.5)(x)+(15.5)(dx)/(dt)+(20) is x=2e^{(-t-c_{1})/(31)}+400