Error checking in DE solution

N

Nazariy

Junior Member
Joined
Jan 21, 2014
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124
Hello,
Can someone please check the following working, as I am not getting the required result unless I multiply my final answer by minus one.

\(\displaystyle \displaystyle{xv\frac{dv}{dx}+2v^2-2=0}\)

\(\displaystyle \displaystyle{\frac{dv}{dx}=\frac{2(1-v^2)}{xv}}\)

\(\displaystyle \displaystyle{\int \frac{v}{1-v^2} dv = 2\ln |x| + c_1}\)

\(\displaystyle \displaystyle{\frac{1}{2} \int \frac{1}{1-v} dv - \frac{1}{2} \int \frac{1}{1+v} dv = -\frac{1}{2} \ln |1-v| - \frac{1}{2} \ln |1+v| + c_2}\)
\(\displaystyle \displaystyle{4\ln|x| = -\ln(|1-v||1+v|)+c_3}\)
​
Now I have an issue with modulus, is there a convention stating that we should take positive values of modulus, just like with \(\displaystyle \sqrt{x^2}=|x|\), or do I have to make an assumption here?
Clearly for \(\displaystyle 4\ln|x| = \ln|x|^4\) so just \(\displaystyle \ln x^4\) because the value will be positive, what about the other logarithm, I think that is where I make an error:

\(\displaystyle \displaystyle{e^{c_3-\ln(|1-v||1+v|)}=x^4}\)

\(\displaystyle \displaystyle{e^{c_3} * e^{\ln (|1-v||1+v|)^-1} = x^4}\)

​
Now I am dropping modulus because I assume that v is positive and less than one.... Is this even correct? This is the dodgy step...
\(\displaystyle Let \ c=e^{c_3}, then \ c*\frac{1}{1-v^2}=x^4\)
\(\displaystyle Since \ y=xv, v =y/x, so \ c=x^4(1-v^2), therefore \ c=x^4-y^2x^2\)
\(\displaystyle c=x^2(x^2-y^2)\)
​
I am supposed to get \(\displaystyle c=x^2(y^2-x^2)\)
There is an easy way around this, I just multiply everything by -1 and I get that result, calling my new negative constant just c. So unless I made an error, this is how I could show that. Is that correct, or have I made an error? Also, how do I treat that modulus in ln?

Thanks
 
Nazariy said:
Hello,
Can someone please check the following working, as I am not getting the required result unless I multiply my final answer by minus one.

\(\displaystyle \displaystyle{xv\frac{dv}{dx}+2v^2-2=0}\)

\(\displaystyle \displaystyle{\frac{dv}{dx}=\frac{2(1-v^2)}{xv}}\)

\(\displaystyle \displaystyle{\int \frac{v}{1-v^2} dv = 2\ln |x| + c_1}\)

\(\displaystyle \displaystyle{\frac{1}{2} \int \frac{1}{1-v} dv - \frac{1}{2} \int \frac{1}{1+v} dv = -\frac{1}{2} \ln |1-v| - \frac{1}{2} \ln |1+v| + c_2}\)
\(\displaystyle \displaystyle{4\ln|x| = -\ln(|1-v||1+v|)+c_3}\)
​
Now I have an issue with modulus, is there a convention stating that we should take positive values of modulus, just like with \(\displaystyle \sqrt{x^2}=|x|\), or do I have to make an assumption here?
Clearly for \(\displaystyle 4\ln|x| = \ln|x|^4\) so just \(\displaystyle \ln x^4\) because the value will be positive, what about the other logarithm, I think that is where I make an error:

\(\displaystyle \displaystyle{e^{c_3-\ln(|1-v||1+v|)}=x^4}\)

\(\displaystyle \displaystyle{e^{c_3} * e^{\ln (|1-v||1+v|)^-1} = x^4}\)

​
Now I am dropping modulus because I assume that v is positive and less than one.... Is this even correct? This is the dodgy step...
\(\displaystyle Let \ c=e^{c_3}, then \ c*\frac{1}{1-v^2}=x^4\)
\(\displaystyle Since \ y=xv, v =y/x, so \ c=x^4(1-v^2), therefore \ c=x^4-y^2x^2\)
\(\displaystyle c=x^2(x^2-y^2)\)
​
I am supposed to get \(\displaystyle c=x^2(y^2-x^2)\)
There is an easy way around this, I just multiply everything by -1 and I get that result, calling my new negative constant just c. So unless I made an error, this is how I could show that. Is that correct, or have I made an error? Also, how do I treat that modulus in ln?

Thanks
Click to expand...
Initially you would need to keep the modulus so that your answer ends up as
\(\displaystyle \pm\, c\, =\, x^2\, (y^2\, -\, x^2)\)
But, as you mentioned, \(\displaystyle \pm\, c\) is just some constant so let the new c = \(\displaystyle \mp\, c\) and
\(\displaystyle c\, =\, x^2\, (x^2\, -\, y^2)\)

In this particular case it turned out that it didn't make a difference about keeping the modulus but in other cases it may.
 
Ishuda said:
Initially you would need to keep the modulus so that your answer ends up as
\(\displaystyle \pm\, c\, =\, x^2\, (y^2\, -\, x^2)\)
But, as you mentioned, \(\displaystyle \pm\, c\) is just some constant so let the new c = \(\displaystyle \mp\, c\) and
\(\displaystyle c\, =\, x^2\, (x^2\, -\, y^2)\)

In this particular case it turned out that it didn't make a difference about keeping the modulus but in other cases it may.
Click to expand...

How do I arrive at \(\displaystyle \pm c\)? Why did it not make a difference, because of c?
 
Nazariy said:
How do I arrive at \(\displaystyle \pm c\)? Why did it not make a difference, because of c?
Click to expand...
The short answers are 'by keeping the modulus up to that point' and 'yes'.

For the longer answer we have at this point in your solution
\(\displaystyle \displaystyle{e^{c_3} * e^{\ln (|1-v||1+v|)^-1} = x^4}\)
we simplify and get
c4 |1 - v2|-1 = x4
where
c4 = ec3
or
c4 = x2 |x2 - y2|.
If v<1 (x>y) then we have the + and if v\(\displaystyle \ge\)1 (x\(\displaystyle \le\)y) we have the -, thus
c4 = \(\displaystyle \pm\)x2 (x2 - y2)
or multiplying through by \(\displaystyle \pm\),
\(\displaystyle \pm\)c4 = x2 (x2 - y2)
depending on the relative sizes of x and y. Since (c1, c2, c3, and) c4 is an arbitrary constant, choose a different arbitrary constant c=\(\displaystyle \mp\)c4 to get the book answer.

EDIT:Fix dumb \(\displaystyle \le\) misteak - unfortunately still visible below.
 
Last edited:
Ishuda said:
The short answers are 'by keeping the modulus up to that point' and 'yes'.

For the longer answer we have at this point in your solution
\(\displaystyle \displaystyle{e^{c_3} * e^{\ln (|1-v||1+v|)^-1} = x^4}\)
we simplify and get
c4 |1 - v2|-1 = x4
where
c4 = ec3
or
c4 = x2 |x2 - y2|.
If v<1 (x>y) then we have the + and if v\(\displaystyle \le\)1 (x\(\displaystyle \le\)y) we have the -, thus
c4 = \(\displaystyle \pm\)x2 (x2 - y2)
or multiplying through by \(\displaystyle \pm\),
\(\displaystyle \pm\)c4 = x2 (x2 - y2)
depending on the relative sizes of x and y. Since (c1, c2, c3, and) c4 is an arbitrary constant, choose a different arbitrary constant c=\(\displaystyle \mp\)c4 to get the book answer.
Click to expand...

Love this! This is what I am always looking for in answers, perfect clarity, now I have come to understand what I did not before. This is perfect, thank you.
 
Nazariy said:
Hello,
Can someone please check the following working, as I am not getting the required result unless I multiply my final answer by minus one.

\(\displaystyle \displaystyle{xv\frac{dv}{dx}+2v^2-2=0}\)

\(\displaystyle \displaystyle{\frac{dv}{dx}=\frac{2(1-v^2)}{xv}}\)

\(\displaystyle \displaystyle{\int \frac{v}{1-v^2} dv = 2\ln |x| + c_1}\)

\(\displaystyle \displaystyle{\frac{1}{2} \int \frac{1}{1-v} dv - \frac{1}{2} \int \frac{1}{1+v} dv = -\frac{1}{2} \ln |1-v| - \frac{1}{2} \ln |1+v| + c_2}\)
\(\displaystyle \displaystyle{4\ln|x| = -\ln(|1-v||1+v|)+c_3}\)
​
Now I have an issue with modulus, is there a convention stating that we should take positive values of modulus, just like with \(\displaystyle \sqrt{x^2}=|x|\), or do I have to make an assumption here?
Clearly for \(\displaystyle 4\ln|x| = \ln|x|^4\) so just \(\displaystyle \ln x^4\) because the value will be positive, what about the other logarithm, I think that is where I make an error:

\(\displaystyle \displaystyle{e^{c_3-\ln(|1-v||1+v|)}=x^4}\)

\(\displaystyle \displaystyle{e^{c_3} * e^{\ln (|1-v||1+v|)^-1} = x^4}\)

​
Now I am dropping modulus because I assume that v is positive and less than one.... Is this even correct? This is the dodgy step...
\(\displaystyle Let \ c=e^{c_3}, then \ c*\frac{1}{1-v^2}=x^4\)
\(\displaystyle Since \ y=xv, v =y/x, so \ c=x^4(1-v^2), therefore \ c=x^4-y^2x^2\)
\(\displaystyle c=x^2(x^2-y^2)\)
​
I am supposed to get \(\displaystyle c=x^2(y^2-x^2)\)
There is an easy way around this, I just multiply everything by -1 and I get that result, calling my new negative constant just c. So unless I made an error, this is how I could show that. Is that correct, or have I made an error? Also, how do I treat that modulus in ln?

Thanks
Click to expand...
Why are you using partial fractions? The differential of the denominator is exactly the numerator up to a constant.
 
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