- #1
Manni
- 42
- 0
I'm a bit confused as to what I do next with this problem.
Consider the initial value problem [itex]\frac{dy}{dx}[/itex] = a(y-b) where y(0) = y0
With a > 0, b = 0, it represents exponential growth, while a < 0, b = 0 gives exponential decay. With y = T, a = -k, k = TA and y0=T0 it gives Newton's Law of Temperature Change,
[itex]\frac{dy}{dx}[/itex] = -k(T - TA)
Solve the given model and give a qualitative sketch for each of the cases 1) a > 0 and 2) a < 0. You may assume b ≥ 0. Include y0= b as one typical solution for each.
So what I did was I applied the integral to both sides of [itex]\frac{dy}{dx}[/itex] = -k(T - TA) yielding,
1 = -k ∫ (T - TA)
But I don't know how to integrate TA
Consider the initial value problem [itex]\frac{dy}{dx}[/itex] = a(y-b) where y(0) = y0
With a > 0, b = 0, it represents exponential growth, while a < 0, b = 0 gives exponential decay. With y = T, a = -k, k = TA and y0=T0 it gives Newton's Law of Temperature Change,
[itex]\frac{dy}{dx}[/itex] = -k(T - TA)
Solve the given model and give a qualitative sketch for each of the cases 1) a > 0 and 2) a < 0. You may assume b ≥ 0. Include y0= b as one typical solution for each.
So what I did was I applied the integral to both sides of [itex]\frac{dy}{dx}[/itex] = -k(T - TA) yielding,
1 = -k ∫ (T - TA)
But I don't know how to integrate TA
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