# Differential equation (substitution)

1. Apr 18, 2014

### delsoo

1. The problem statement, all variables and given/known data

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]

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 6, 2017
2. Apr 18, 2014

### Simon Bridge

"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.

3. Apr 18, 2014

### LCKurtz

And type the equations here instead of graphics on alternate sites.

4. Apr 19, 2014

### delsoo

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!

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Last edited: Apr 19, 2014
5. Apr 19, 2014

### Simon Bridge

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.