Question about integral differential equation

[EllieJones]

I have been working on this problem for quite some time. I found an answer eventually, but Pearson insists that my solution is wrong. I tried several times, but I still came up with the same answer. Can somebody help me?

 

[HallsofIvy]

After all that excelent math, you have a silly little miscopying!

Yes, integrating gives \(\displaystyle \frac{1}{y}= \frac{1}{36(9x^2+ 2)^2}+ C\).

When x= 0, y= 2 so \(\displaystyle \frac{1}{2}= \frac{1}{36(9(0^2)+ 2)^2}+ C\).
But then you have \(\displaystyle \frac{1}{2}= \frac{1}{36(0+ 1)^2}+ C\) while you should have \(\displaystyle \frac{1}{2}= \frac{1}{36(2)^2}+ C\).

\(\displaystyle \frac{1}{2}= \frac{1}{36(4)}+ C= \frac{1}{144}+ C\).
\(\displaystyle C= \frac{1}{2}- \frac{1}{144}= \frac{72}{144}- \frac{1}{144}= \frac{71}{144}\).

 

[EllieJones]

After all that excelent math, you have a silly little miscopying!

Yes, integrating gives \(\displaystyle \frac{1}{y}= \frac{1}{36(9x^2+ 2)^2}+ C\).

When x= 0, y= 2 so \(\displaystyle \frac{1}{2}= \frac{1}{36(9(0^2)+ 2)^2}+ C\).
But then you have \(\displaystyle \frac{1}{2}= \frac{1}{36(0+ 1)^2}+ C\) while you should have \(\displaystyle \frac{1}{2}= \frac{1}{36(2)^2}+ C\).

\(\displaystyle \frac{1}{2}= \frac{1}{36(4)}+ C= \frac{1}{144}+ C\).
\(\displaystyle C= \frac{1}{2}- \frac{1}{144}= \frac{72}{144}- \frac{1}{144}= \frac{71}{144}\).

Oh my god, I'm such an idiot! The thing is I don't really have a good math foundation so I always make these little mistakes. Thank you!

 

[topsquark]

Oh my god, I'm such an idiot! The thing is I don't really have a good math foundation so I always make these little mistakes. Thank you!

The worst problem in solving a question is finding the little mistakes. Nothing to be embarrassed about.

-Dan