Find the Differential Equation (Integrating Factor Problem)

Question

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.

Answer 1

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 ?

Answer 2

$$y'x+y=3xy$$ observe that $$(xy)'=3xy$$ Integrate $$\int \frac {d(xy)}{xy}=\int 3 dx$$ $$\ln|xy|=3x+K$$ $$\ln|y|=3x+K-\ln|x|$$ $$y(x)=........$$