Differencial Equation

[RoxieApoxie]

Having some trouble trying to decipher what this is exactly, I tried to solve it (I'll explain after the question)

Suppose that in a certain company, the relationship between the price per unit p of its product and the weekly sales volume y, in thousands of dollars, is given by

dy/dp= -2/5(y/p+8)

Solve the differencial equation when y(24)=8

The professor is giving us the answer of -2.5ln(y)=ln(p+8)-8.66

My first reaction was "WHAT? Where did this ln come from? I did some research to try and solve it, and this is where I got stuck and/or completely lost:

5(p+8)dy=-2y(dx)

dx/5(p+8) - dy/-2y

Antiderivative: dx/5(p+8) - Antiderivative dy/-2y = C

ln(p+8) ....

uhh, LOL, I'm completely lost on this one. I'm not sure how to even set up this equation, but I tried (which is probably completely wrong anyway!) Thanks again for your help

 

[Deleted member 4993]

Having some trouble trying to decipher what this is exactly, I tried to solve it (I'll explain after the question)

Suppose that in a certain company, the relationship between the price per unit p of its product and the weekly sales volume y, in thousands of dollars, is given by

dy/dp= -2/5(y/p+8)

Solve the differencial equation when y(24)=8

The professor is giving us the answer of -2.5ln(y)=ln(p+8)-8.66

My first reaction was "WHAT? Where did this ln come from? I did some research to try and solve it, and this is where I got stuck and/or completely lost:

5(p+8)dy=-2y(dx) ................................. where did 'x' come from?

dx/5(p+8) - dy/-2y

Antiderivative: dx/5(p+8) - Antiderivative dy/-2y = C

ln(p+8) ....

uhh, LOL, I'm completely lost on this one. I'm not sure how to even set up this equation, but I tried (which is probably completely wrong anyway!) Thanks again for your help

dy/dp= -2/5(y/p+8)

\(\displaystyle -\frac{5}{2}\cdot \frac{dy}{y} \ = \ \frac{dp}{p+8}\)

Now integrate both sides.......

 

[soroban]

Hello, RoxieApoxie!

\(\displaystyle \frac{dy}{dp} \:=\: -\frac{2}{5}\,\frac{y}{p+8}\)

\(\displaystyle \text{Solve the differencial equation if: }\,y(24)\,=\,8\)

\(\displaystyle \text{Separate variables: }\:5\,\frac{dy}{y} \:=\:-2\,\frac{dp}{p+8}\)

\(\displaystyle \text{Integrate: }\:5\int\frac{dy}{y} \:=\:-2\int\frac{dp}{p+8}\)

. . . . . . . . \(\displaystyle 5\ln(y) \:=\:- 2\ln(p+8) + C\) .[1]

\(\displaystyle \begin{Bmatrix}p=24 \\ y=8\end{Bmatrix} \;\; 5\ln(8) \:=\:-2\ln(32) + C\)

. . . . . . . \(\displaystyle 5\ln(2^3) \:=\:-2\ln(2^5) + C\)

. . . . . . .\(\displaystyle 15\ln(2) \:=\:-10\ln(2) + C\)

. . . . . . .\(\displaystyle 25\ln(2) \:=\:C\)

. = . . . . . . . . \(\displaystyle C \:=\:\ln(2^{25})\;\)


Substitute into [1]:

. . \(\displaystyle 5\ln(y) \;=\;-2\ln(p+8) + \ln\left(2^{25}\right)\)

. . \(\displaystyle \ln(y^5) \;=\;\ln(p+8)^{-2} + \ln\left(2^{25}\right)\)

. . \(\displaystyle \ln(y^5) \;=\;\ln\left[2^{25}(p+8)^{-2}\right]\)

. = . . \(\displaystyle y^5 \;=\;2^{25}(p+8)^{-2}\)

. = . . .\(\displaystyle y \;=\;2^5(p+8)^{-\frac{2}{5}}\)

. = . . .\(\displaystyle y \;=\;\frac{32}{(p+8)^{\frac{2}{5}}}\)

 

[mmm4444bot]

dy/dp= -2/5(y/[p + 8])

Solve the [differential] equation when y(24) = 8

The professor [gave] us the answer of -2.5 ln(y) = ln(p + 8) - 8.66

I do not agree with your professor. :shock:

 

[RoxieApoxie]

Thank you so much for your help everyone! I hadn't realized I had to separate the variables like that.

mmmbot, is my professor as lost as I am/was?