Separation of Variables Math Example 4

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Example 4

medium
Solve Newton's cooling: dT/dt=−0.1(T−20)dT/dt = -0.1(T-20), T(0)=80T(0)=80.

Solution

  1. 1
    Let u=T−20u = T-20: du/dt=−0.1udu/dt = -0.1u. Solution: u=Ae−0.1tu = Ae^{-0.1t}.
  2. 2
    u(0)=60⇒A=60u(0) = 60 \Rightarrow A=60. T=20+60e−0.1tT = 20+60e^{-0.1t}.

Answer

T(t)=20+60e−0.1tT(t) = 20 + 60e^{-0.1t}
Substituting u=T−Tsu = T-T_s reduces Newton's cooling to exponential decay.

About Separation of Variables

A method for solving first-order DEs of the form dydx=f(x)⋅g(y)\frac{dy}{dx} = f(x) \cdot g(y): rearrange to dyg(y)=f(x) dx\frac{dy}{g(y)} = f(x)\,dx, then integrate both sides.

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More Separation of Variables Examples

Example 1 easy

Solve [formula] with [formula].

Example 2 hard

Solve the logistic DE [formula] with [formula].

Example 3 easy

Solve [formula] with [formula].

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