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C4 integration question

jasminetwine

Hi guys,

Is anyone able to help me with this integration question from the Heinemann textbook? :smile:

I've managed to get up to here so far:

Attachment not found

But I don't know where to go now :frown:

(The final answer is y^2 = 8x/(x+2))

Thanks in advance :smile:

image.jpeg5.7KB

Reply 1

Zacken

Original post by jasminetwine

Hi guys,

Is anyone able to help me with this integration question from the Heinemann textbook? :smile:

I've managed to get up to here so far:

Attachment not found

But I don't know where to go now :frown:

(The final answer is y^2 = 8x/(x+2))

Thanks in advance :smile:

I'm assuming you have

\ln y = \frac{1}{2} \ln |x| - \frac{1}{2} \ln |x+2| + c = \frac{1}{2}(\ln |x| - \ln |x+2|) + c

In which case, this gets you

\displaystyle \ln y = \frac{1}{2}\ln \left|\frac{Ax}{x+2} \right|= \ln \sqrt{\frac{Ax}{x+2}}

for some arbitrary constant

A

, using the power rule for logarithms and "cancelling" the logs gets you

\displaystyle y = \sqrt{\frac{Ax}{x+2}}

and can you then use the given information to find

A

?

Alternatively, you could have done:

Unparseable latex formula:

\ln y = \frac{1}{2} \left|\frac{Ax}{x+2}right| \Rightarrow 2 \ln y = ln \frac{Ax}{x+2} \Rightarrow \ln y^2 = \ln \frac{Ax}{x+2}

and using the power rule on

2\ln y

which is more elegant in my opinion.

(edited 8 years ago)

Reply 2

jasminetwine

Original post by Zacken

I'm assuming you have

\ln y = \frac{1}{2} \ln |x| - \frac{1}{2} \ln |x+2| + c = \frac{1}{2}(\ln |x| - \ln |x+2|) + c

In which case, this gets you

\displaystyle \ln y = \frac{1}{2}\ln \left|\frac{Ax}{x+2} \right|= \ln \sqrt{\frac{Ax}{x+2}}

for some arbitrary constant

A

, using the power rule for logarithms and "cancelling" the logs gets you

\displaystyle y = \sqrt{\frac{Ax}{x+2}}

and can you then use the given information to find

A

?

Alternatively, you could have done:

Unparseable latex formula:

\ln y = \frac{1}{2} \left|\frac{Ax}{x+2}right| \Rightarrow 2 \ln y = ln \frac{Ax}{x+2} \Rightarrow \ln y^2 = \ln \frac{Ax}{x+2}

and using the power rule on

2\ln y

which is more elegant in my opinion.

How did you get from...

To...

Attachment not found

Thank you! :smile:

image.jpeg24.5KB

Reply 3

the bear

Original post by jasminetwine

How did you get from...

To...

Attachment not found

Thank you! :smile:

you can write c as 0.5lnA without any loss of generality...

Reply 4

Zacken

Original post by jasminetwine

How did you get from...

To...

Attachment not found

Thank you! :smile:

It's an arbitrary constant, so I can do whatever I'd like with it - but if it makes you happy, we can do this:

\ln y = \frac{1}{2} \ln x - \frac{1}{2} \ln |x+2| + \frac{1}{2} \ln e^{2c}

since

c = \ln e^c \Rightarrow c = \frac{1}{2} \ln e^{2c}

So that I get

2\ln y = \ln \frac{e^{2c}x}{x+2}

but then you can just say

A=e^{2c}

without needing to bother for the sake of simplicity.

So we get

\ln y^2 = \ln \frac{Ax}{x+2}

once again, although you could have just as well left it in the form above if you wanted.

Reply 5

nerak99

Original post by jasminetwine

How did you get from...

To...

Attachment not found

Thank you! :smile:

Call c, 0.5 ln A and then take out the half as a common factor. Inside the bracket use the rules of logs (log a + log b = log ab and log a - log b = log a/b) to collect together the log A, log x and - log |x+2| into one log.

The trick of changing c to log k is very common in doing the algebra in tidying up solutions of DEs