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This is #11 of Section 1.5 of the book DEALA, 3E. So, we have only learned how to solve by using the integrating factor. BUT, there is a factor of 'y' on the right-hand side. What is going on?

Find the differential:

$x\frac{dy}{dx} + y = 3xy$, y(1) = 0

Add: The solution I looked at after reviewing comments:

$x\frac{dy}{dx} = 3xy - y$

$x\frac{dy}{dx} = (3x - 1)y$

$\frac{x}{3x-1}\frac{1}{dx} = \frac{y}{dy}$

$\int\frac{3x-1}{x}dx = \int\frac{1}{y}dy$

$\int3dx - \int\frac{1}{x}dx = \int\frac{1}{y}dy$

$ln|y| = 3x - ln|x| + C$

But, the solution in the back is y is parallel to 0. I think it's a misprint.

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  • $\begingroup$ This is also eqiuvalent to $xy'+(1-3x)y=0$ $\endgroup$ – Paul Feb 8 '18 at 19:34
  • $\begingroup$ Thank you for the hint. I see how to solve it now. $\endgroup$ – gm9089zo Feb 8 '18 at 19:38
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No need for an integrating factor, it is variable seperable ! \begin{eqnarray*} x \frac{dy}{dx}=(3x-1)y. \end{eqnarray*} Should be easy from here ?

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  • $\begingroup$ Yes, thank you. I did not realize that. I have the solution now. $\endgroup$ – gm9089zo Feb 8 '18 at 19:51
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$$y'x+y=3xy$$ observe that $$(xy)'=3xy$$ Integrate $$\int \frac {d(xy)}{xy}=\int 3 dx$$ $$\ln|xy|=3x+K$$ $$\ln|y|=3x+K-\ln|x|$$ $$y(x)=........$$

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