What is the solution for (dr)/(dt)=-4rt,r=r_{0}(t),t=0 ?

The solution for (dr)/(dt)=-4rt,r=r_{0}(t),t=0 is r=e^{-2t^2+c_{1}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dr)/(dt)=-4rt,r=r_{0}(t),t=0

\frac{d r}{d t} = - 4 r t, r = r_{0} (t), t = 0

r = e^{- 2 t^{2} + c_{1}}

Solution steps

\frac{d r}{d t} = - 4 r t

Solve separable ODE:

r = e^{- 2 t^{2} + c_{1}}

r = e^{- 2 t^{2} + c_{1}}

y^{^{'}} - \frac{x ^{2} y - y}{y + 4} = 0, y (3) = 1

(e^{m} + 1)^{2} e^{- m} d t + (e^{t} + 1)^{3} e^{- t} d m = 0

\frac{d y}{d x} + cos (x) y - cos (x) = 0

\frac{d y}{d x} = \frac{1 + x}{x y ^{5}}

2 x^{2} y^{^{'}} = y^{2}

What is the solution for (dr)/(dt)=-4rt,r=r_{0}(t),t=0 ?
The solution for (dr)/(dt)=-4rt,r=r_{0}(t),t=0 is r=e^{-2t^2+c_{1}}