How does dy/dt=ry end up with the constant multiple on the right?

In summary, the conversation discusses the solution to the differential equation dy/dt = ry, which results in y = e^(rt)+C. However, the book presents a different solution where y = y_0 * e^(rt), with y_0 representing the initial condition y(0). The confusion arises from the inclusion of the constant of integration, which is represented by exp(C) in the first solution and simply replaced by a new constant C in the second solution. The mistake was most likely not including the constant of integration immediately after integrating.
  • #1
219
0
I have, from the textbook:

dy/dt = ry

So I solve and get y = e^(rt)+C

However, the book says that y=y_0 * e^(rt), where y_0 is the answer to the initial condition, y(0).

I don't get it. The + C should be what we use to solve for the initial condition; how does y_0 end up in the same term with the exponential?

Thanks
 
Physics news on Phys.org
  • #2
integrating dy/y=rdt leads to ln(y)=rt+C

this leads to y=exp(rt+C)
but exp(rt+C)=exp(C)*exp(rt)
and you simply replace the constant exp(C) by a new constant C

EDIT: your mistake was most likely that you didn't include the constant of integration immediately after integrating, but after getting rid of the logarithm.
 
  • #3
Yes! I see. Thank you.
 

1. How does the derivative dy/dt end up with a constant multiple on the right?

The derivative dy/dt is the rate of change of a function y with respect to time t. When we apply the product rule of differentiation to a function y = r*t, where r is a constant, the derivative dy/dt becomes r, which is a constant multiple on the right side. This is because the derivative of a constant is always 0.

2. Why is the constant multiple on the right side of the derivative equation?

The constant multiple on the right side of the derivative equation is a result of the product rule, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. In the case of dy/dt = r, the derivative of the first function (r) is 0, so only the second function (t) remains, resulting in the constant multiple on the right side.

3. Does the constant multiple on the right side affect the value of the derivative?

No, the constant multiple on the right side does not affect the value of the derivative. The value of the derivative is still equal to the rate of change of the function y with respect to time t, which is r in this case. The constant multiple on the right side is simply a representation of the constant rate of change of the function.

4. Can the constant multiple on the right side vary?

No, the constant multiple on the right side cannot vary. As mentioned before, the constant multiple represents the constant rate of change of the function and cannot change. However, if the constant r in the function y = r*t is changed, the value of the derivative will also change accordingly.

5. How does understanding the constant multiple on the right side help in solving problems?

Understanding the constant multiple on the right side helps in solving problems because it allows us to easily differentiate functions that involve a constant rate of change, such as in the case of exponential growth or decay. It also helps us to understand the relationship between the original function and its derivative, and how changes in the constant can affect the derivative and vice versa.

Suggested for: How does dy/dt=ry end up with the constant multiple on the right?

Back
Top