problem with last step in differential equation and solution

mathstresser

Junior Member
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Jan 28, 2006
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Find the general solution of the indicated differential equation.

\(\displaystyle \L\\ xy'\, -\, y\, =\, 2x^2 y\)

*That is xy-prime - y =...

So, I get...

\(\displaystyle \L\\ x\, \frac{dy}{dx}\, =\, 2x^2 y\,dx\, +\, y\, dx\)

\(\displaystyle \L\\ x\,dy\,=\, dx(2x^2 y\, +\, y)\)

\(\displaystyle \L\\ x\,dy\, =\, dx(y)(1\,+\,2x^2)\)

\(\displaystyle \L\\ \frac{dy}{y}\,=\, \left(\,\frac{1\,+\,2x^2}{x}\,\right)\,dx\)

The last part is where I have problems. I can do the first integral, but not the second.

ln y = ???
 
\(\displaystyle \L \frac{1 + 2x^2}{x} = \frac{1}{x} + 2x\)

\(\displaystyle \L \int \frac{1}{x} + 2x dx\) = ?
 
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