Why do partial fractions need to be simplified to 'lowest terms'

[ausmathgenius420]

Hi there,

The textbook question I am working on requires me to solve;
[math]\frac{dx}{dy}=\frac{-50}{(5-y)(10-y)}[/math]
Since my solution was wrong, I plugged the equation into wolfram, and they make this step of working.

Express as a ratio of expanded polynomials in lowest terms:
[math]\frac{-50}{(y-5)(y-10)}[/math]
Is this a necessary step to integrating with partial fractions? I can't see any other errors in my working.

 

[BigBeachBanana]

Hi there,

The textbook question I am working on requires me to solve;
[math]\frac{dx}{dy}=\frac{-50}{(5-y)(10-y)}[/math]
Since my solution was wrong, I plugged the equation into wolfram, and they make this step of working.

Express as a ratio of expanded polynomials in lowest terms:
[math]\frac{-50}{(y-5)(y-10)}[/math]
Is this a necessary step to integrating with partial fractions? I can't see any other errors in my working.

The step is not necessary. Please share your work, otherwise, we can't tell you why your work is wrong.

Note: It's possible to solve without partial fractions. It can be solved by doing substitution twice.

 

[ausmathgenius420]

The step is not necessary. Please share your work, otherwise, we can't tell you why your work is wrong.

Note: It's possible to solve without partial fractions. It can be solved by doing substitution twice.

Attached is my working. I didn't bother putting y in terms of x as something above has already gone wrong... the answer sheet has e^-0.1x. Can you see what I've done wrong?

 

[BigBeachBanana]

View attachment 33024
Attached is my working. I didn't bother putting y in terms of x as something above has already gone wrong... the answer sheet has e^-0.1x. Can you see what I've done wrong?

The mistake started at the second line.

[math]\frac{dy}{dx}=-\frac{(y-10)(y-5)}{50}\\ \frac{dx}{dy}=-\frac{50}{(y-10)(y-5)}\\ \int dx = \int -\frac{50}{(y-10)(y-5)}dy [/math]
Continue from here...

 

[ausmathgenius420]

The mistake started at the second line.

[math]\frac{dy}{dx}=-\frac{(y-10)(y-5)}{50}\\ \frac{dx}{dy}=-\frac{50}{(y-10)(y-5)}\\ \int dx = \int -\frac{50}{(y-10)(y-5)}dy [/math]
Continue from here...

Not sure why you changed the order of the variables... Is this not correct?

[math]\frac{dy}{dx}=-(1-\frac{y}{5})(1-\frac{y}{10})[/math]
[math]\frac{dy}{dx}=-(\frac{5-y}{5})(\frac{10-y}{10})[/math]
[math]\frac{dy}{dx}=-\frac{(5-y)(10-y)}{50}[/math]

 

[BigBeachBanana]

Not sure why you changed the order of the variables... Is this not correct?

[math]\frac{dy}{dx}=-(1-\frac{y}{5})(1-\frac{y}{10})[/math]
[math]\frac{dy}{dx}=-(\frac{5-y}{5})(\frac{10-y}{10})[/math]
[math]\frac{dy}{dx}=-\frac{(5-y)(10-y)}{50}[/math]

My mistake, you're correct that they're equivalent. I didn't work the problem all the way through. I believe your mistake started after the integration. Give me some time to work it out. Thanks.

 

[ausmathgenius420]

My mistake, you're correct that they're equivalent. I didn't work the problem all the way through. I believe your mistake started after the integration. Give me some time to work it out. Thanks.

Thanks for checking, attached is the textbook solution.

 

[BigBeachBanana]

The only mistake I see in your work is when solving for A. [imath]\frac{15}{10}=\frac{3}{2}[/imath], not [imath]\frac{2}{3}[/imath]

If you were to continue, you'd have the correct answer. Getting [imath]e^{0.1x}[/imath] isn't necessarily wrong. Just needed an extra algebra step to match their solution.
Let's pick up from here:

Solve for y:

[math]Ae^{0.1x}=\frac{y-5}{y-10}\\ Ae^{0.1x}y-10Ae^{0.1x}=y-5\\ y=\frac{-5+10Ae^{0.1x}}{Ae^{0.1x}-1}\\ y=\frac{-5+10(3/2)e^{0.1x}}{(3/2)e^{0.1x}-1}\\ \text{Simplify by multiply top and bottom by 2:}\\ y=\frac{-10+30e^{0.1x}}{3e^{0.1x}-2}\\[/math][math]\text{To match their solution, divide top and bottom by }e^{0.1x}:\\ y=\frac{-10e^{-0.1x}+30}{3-2e^{-0.1x}}\\[/math]

 

[ausmathgenius420]

The only mistake I see in your work is when solving for A. [imath]\frac{15}{10}=\frac{3}{2}[/imath], not [imath]\frac{2}{3}[/imath]
View attachment 33030
If you were to continue, you'd have the correct answer. Getting [imath]e^{0.1x}[/imath] isn't necessarily wrong. Just needed an extra algebra step to match their solution.
Let's pick up from here:
View attachment 33029
Solve for y:

[math]Ae^{0.1x}=\frac{y-5}{y-10}\\ Ae^{0.1x}y-10Ae^{0.1x}=y-5\\ y=\frac{-5+10Ae^{0.1x}}{Ae^{0.1x}-1}\\ y=\frac{-5+10(3/2)e^{0.1x}}{(3/2)e^{0.1x}-1}\\ \text{Simplify by multiply top and bottom by 2:}\\ \tag{This is acceptable as well}y=\frac{-10+30e^{0.1x}}{3e^{0.1x}-2}\\[/math][math]\text{To match their solution, divide top and bottom by }e^{0.1x}:\\ y=\frac{-10e^{-0.1x}+30}{3-2e^{-0.1x}}\\[/math]

I understand now, thanks so much for your help!