What is the solution for (dh)/(dt)=-0.03sqrt(h-19) ?

The solution for (dh)/(dt)=-0.03sqrt(h-19) is h=(9t^2}{40000}-(3c_{1}t)/(200)+\frac{c_{1}^2)/4+19

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dh)/(dt)=-0.03sqrt(h-19)

\frac{d h}{d t} = - 0.03 h - 19

h = \frac{9 t ^{2}}{40000} - \frac{3 c _{1} t}{200} + \frac{c _{1}^{2}}{4} + 19

Solution steps

\frac{d h}{d t} = - 0.03 h - 19

Solve separable ODE:

h = \frac{9 t ^{2}}{40000} - \frac{3 c _{1} t}{200} + \frac{c _{1}^{2}}{4} + 19

h = \frac{9 t ^{2}}{40000} - \frac{3 c _{1} t}{200} + \frac{c _{1}^{2}}{4} + 19

\frac{d y}{d x} = 8 (2 - y)

y^{^{'}} = 4 x y - 2 x, y (0) = 4

(x + 1) y^{^{'}} + y = 0

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

\frac{d y}{d x} = 6 x y^{2}

What is the solution for (dh)/(dt)=-0.03sqrt(h-19) ?
The solution for (dh)/(dt)=-0.03sqrt(h-19) is h=(9t^2}{40000}-(3c_{1}t)/(200)+\frac{c_{1}^2)/4+19