"rate of change proportional to the diffrence between itself and a fixed value dy/dx=k(y−a) So I have a question regarding exponential growth/decay problem formulated in my book. In my book they say: ""Sometimes an exponential growth or decay problem will involve a quantity that changes at a rate proportional to the difference between itself and a fixed point: dy/dx=k(y−a) In this case, the change of dependent variable u(t)=y(t)−a should be used to convert the differential equation to the standard form. Observe that u(t) changes at the same rate as $y(t)(ie,\frac{du {dy}=dy/dt) so it satisfies du/dt=ku"" I am having a hard time grasping what they mean. And I would hope for somebody to expand on this notion. More specifically how does dy/dx=k(y−a) relate to u(t)=y(t)−a?"

Keenan Conway

Keenan Conway

Answered question

2022-09-24

rate of change proportional to the diffrence between itself and a fixed value d y d x = k ( y a )
So I have a question regarding exponential growth/decay problem formulated in my book.
In my book they say:
"Sometimes an exponential growth or decay problem will involve a quantity that changes at a rate proportional to the difference between itself and a fixed point:
d y d x = k ( y a )
In this case, the change of dependent variable u(t)=y(t)−a should be used to convert the differential equation to the standard form. Observe that u(t) changes at the same rate as $y(t)(ie,\frac{du}{dy}=\frac{dy}{dt}) so it satisfies
d u d t = k u
I am having a hard time grasping what they mean. And I would hope for somebody to expand on this notion.
More specifically how does d y d x = k ( y a ) relate to u ( t ) = y ( t ) a?

Answer & Explanation

booyo

booyo

Beginner2022-09-25Added 6 answers

since we have
u ( t ) = y ( t ) a
we get by differentiating with respect to t
d u d t = d y d t
(since a is const.) so we get
d u ( t ) d t = k u ( t )

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