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#### alane1994

##### Active member

- Oct 16, 2012

- 126

Ok, I have a practice exam... My professor gave out a copy with worked out examples. There is one where I don't get his logic at all. I was wondering if you guys could explain it to me?

\(\displaystyle (\frac{1}{t}+2y^2t)dt+(2yt^2-\cos(y))dy=0\)

First, he put \(\text{Assume t>0}\)

???

\(\displaystyle \text{Since} \frac{\partial}{\partial y}(\frac{1}{t}+2y^2t)=4yt=\frac{\partial}{\partial t}(2yt^2-\cos(y))\)

\(\displaystyle F(t,y)=\int (\frac{1}{t}+2y^2t)dt=\ln(t)+y^2t^2+f(y)\)

\(\displaystyle ~~~~~~~~~~~=\int (2yt^2-\cos(y))dy=y^2ts-\sin(y)+g(t)\)

From those (not sure how to do two lines into right brace).

\(\ln(t)+t^2y^2-\sin(y)=C\)

\(y(1)=\pi\)

\(0+\pi^2-0=C\)

\(\therefore \ln(t)+t^2+y^2-\sin(y)=\pi^2\)

\(\displaystyle (\frac{1}{t}+2y^2t)dt+(2yt^2-\cos(y))dy=0\)

First, he put \(\text{Assume t>0}\)

???

\(\displaystyle \text{Since} \frac{\partial}{\partial y}(\frac{1}{t}+2y^2t)=4yt=\frac{\partial}{\partial t}(2yt^2-\cos(y))\)

\(\displaystyle F(t,y)=\int (\frac{1}{t}+2y^2t)dt=\ln(t)+y^2t^2+f(y)\)

\(\displaystyle ~~~~~~~~~~~=\int (2yt^2-\cos(y))dy=y^2ts-\sin(y)+g(t)\)

From those (not sure how to do two lines into right brace).

\(\ln(t)+t^2y^2-\sin(y)=C\)

\(y(1)=\pi\)

\(0+\pi^2-0=C\)

\(\therefore \ln(t)+t^2+y^2-\sin(y)=\pi^2\)

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