# Differential equation (substitution)

## Homework Statement

for this question , i 've got my positive 9 but i got -64 , can anyone tell me which part is wrong?

Question : https://www.flickr.com/photos/123101...3/13907725466/ [Broken]
Wroking : https://www.flickr.com/photos/123101...n/photostream/ [Broken]

## The Attempt at a Solution

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Simon Bridge
Homework Helper
"flickr: page no found"
Note: The way to get the most out of this site is to write about how you are thinking about the problem - then we can focus on why you lost so many marks and fix that for the future.

LCKurtz
Homework Helper
Gold Member
And type the equations here instead of graphics on alternate sites.

sorry, i posted at the wrong section, , but i hope you can look at it and help! here's the question and working attached..hope you can help!
can i do it in this way? please refer to the red circled part

if i cant do it in such way, can you please show me how do u get the ans please? thanks in advance!

#### Attachments

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Last edited:
Simon Bridge
Homework Helper
Using substitution ##y=Vx##, change the DE $$(x^3+xy^2)\frac{dy}{dx}=y^3+yx^2$$... to $$x\frac{dV}{dx}=\frac{-2V}{1+V^2}$$... and show that the solution is $$\ln\left|\frac{xy}{2}\right| = \frac{(x-2y)(x+2y)}{8x^2}$$ is ##x=2## when ##y=1##.

Good grief!

Well the first part comes from making the sub and then a bit of algebra.
You could have cancelled the x^3 out much earlier than you did (like your 3rd line).
But you made it - well done.

For the next bit you basically started with separation of variables: $$\int \frac{dx}{x} =\int \frac{1+V^2}{-2V}dV$$... then you wrote:$$-2\int \frac{\frac{1}{2}dx}{\frac{1}{2}x} =\int \frac{1+V^2}{V}dV$$... which is the bit in red. It's a legal move.
You seem concerned that $$\int \frac{\frac{1}{2}dx}{\frac{1}{2}x}=\ln\left|\frac{1}{2}x\right|+c$$ ##\qquad\qquad## ... instead of ##\ln|x|## but this is fine.

Recall: ##\ln|\frac{1}{2}x|## differs from ##\ln|x|## by a constant that gets hidden in that "+c" on the end.