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.
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:
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
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? \(\displaystyle \displaystyle{\frac{dv}{dx}=\frac{2(1-v^2)}{xv}}\)
\(\displaystyle \displaystyle{\int \frac{v}{1-v^2} dv = 2\ln |x| + c_1}\)
\(\displaystyle \displaystyle{\frac{1}{2} \int \frac{1}{1-v} dv - \frac{1}{2} \int \frac{1}{1+v} dv = -\frac{1}{2} \ln |1-v| - \frac{1}{2} \ln |1+v| + c_2}\)
\(\displaystyle \displaystyle{4\ln|x| = -\ln(|1-v||1+v|)+c_3}\)
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... \(\displaystyle \displaystyle{e^{c_3} * e^{\ln (|1-v||1+v|)^-1} = x^4}\)
\(\displaystyle Let \ c=e^{c_3}, then \ c*\frac{1}{1-v^2}=x^4\)
\(\displaystyle Since \ y=xv, v =y/x, so \ c=x^4(1-v^2), therefore \ c=x^4-y^2x^2\)
\(\displaystyle c=x^2(x^2-y^2)\)
I am supposed to get \(\displaystyle c=x^2(y^2-x^2)\) \(\displaystyle Since \ y=xv, v =y/x, so \ c=x^4(1-v^2), therefore \ c=x^4-y^2x^2\)
\(\displaystyle c=x^2(x^2-y^2)\)
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