What is the solution for y^'=9.8-0.2y,y(0)=100 ?

The solution for y^'=9.8-0.2y,y(0)=100 is y=49+(51)/(e^{0.2t)}

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y^'=9.8-0.2y,y(0)=100

y^{^{'}} = 9.8 - 0.2 y, y (0) = 100

y = 49 + \frac{51}{e ^{0.2 t}}

Solution steps

y^{'} (t) = 9.8 - 0.2 y

Solve linear ODE:

y = 49 + \frac{51}{e ^{0.2 t}}

y = 49 + \frac{51}{e ^{0.2 t}}

n = - 4 \sum \infty (- \frac{4}{9})^{n}

x \to 5 + lim (\frac{x - 5}{∣ x - 5 ∣})

y^{^{'}} = - \frac{y}{2000} 4

(- 4, 0), (0, - 1)

What is the solution for y^'=9.8-0.2y,y(0)=100 ?
The solution for y^'=9.8-0.2y,y(0)=100 is y=49+(51)/(e^{0.2t)}