tonan6e

2022-09-29

Explanation of proof nedeed: Why is ${y}^{\prime}=c\cdot y$ always a exponential growth/decay function?

I know, that a solution of ${y}^{\prime}=c\cdot y$ is $y=a\cdot {e}^{ct}$ and it's clear how to calculate this.

I want to proof, that all solutions of a function describing a change of population, that is proprotional to the population, like ${y}^{\prime}=c\cdot y$ is a function of exponential growth (or decay.

I found a proof:

Let g be an other solution, whilst g is not describing an exponential growth or decay. We show $(\frac{g}{{e}^{ct}}{)}^{\prime}=0$ It's fine to me how they show it's zero. But where does $(\frac{g}{{e}^{ct}}{)}^{\prime}$ come from and why are they using it here, what does it mean?

I know, that a solution of ${y}^{\prime}=c\cdot y$ is $y=a\cdot {e}^{ct}$ and it's clear how to calculate this.

I want to proof, that all solutions of a function describing a change of population, that is proprotional to the population, like ${y}^{\prime}=c\cdot y$ is a function of exponential growth (or decay.

I found a proof:

Let g be an other solution, whilst g is not describing an exponential growth or decay. We show $(\frac{g}{{e}^{ct}}{)}^{\prime}=0$ It's fine to me how they show it's zero. But where does $(\frac{g}{{e}^{ct}}{)}^{\prime}$ come from and why are they using it here, what does it mean?

emarisidie6

Beginner2022-09-30Added 7 answers

It's fine to me how they show it's zero. But where does $(\frac{g}{{e}^{ct}}{)}^{\prime}$ come from and why are they using it here, what does it mean?

I'm not sure what it $g(t)/{e}^{ct}$ means in and of itself. But it a means to an end. (excuse my play on words there).

Our aim is to reveal that a feature satisfying a sure differential equation must take a sure form. The principal equation of this form, and its solution, is the reality that a characteristic whose by-product is zero (on an c language) need to be a consistent (on that interval).

So we prepare dinner up a associated function which, if it have been consistent, would tell us the feature we're originally looking at ought to have the shape we are claiming it to have. Then we take the derivative of the associated function and display it's miles zero.

Because we want to show $g(t)=a{e}^{ct}$ for some a, we construct the quotient $h(t)=g(t)/{e}^{ct}$. This function is constant if and only if g has the form we claim it does. Then we prove that h is constant by showing ${h}^{\prime}(t)=0$.

This is not obvious to anyone seeing it for the first time, but now you know the technique: find something you want to be constant, and prove that it must be constant by showing its derivative is zero.

I'm not sure what it $g(t)/{e}^{ct}$ means in and of itself. But it a means to an end. (excuse my play on words there).

Our aim is to reveal that a feature satisfying a sure differential equation must take a sure form. The principal equation of this form, and its solution, is the reality that a characteristic whose by-product is zero (on an c language) need to be a consistent (on that interval).

So we prepare dinner up a associated function which, if it have been consistent, would tell us the feature we're originally looking at ought to have the shape we are claiming it to have. Then we take the derivative of the associated function and display it's miles zero.

Because we want to show $g(t)=a{e}^{ct}$ for some a, we construct the quotient $h(t)=g(t)/{e}^{ct}$. This function is constant if and only if g has the form we claim it does. Then we prove that h is constant by showing ${h}^{\prime}(t)=0$.

This is not obvious to anyone seeing it for the first time, but now you know the technique: find something you want to be constant, and prove that it must be constant by showing its derivative is zero.

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