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Solve this differential equation using given change of variables $(xy+2xy(\ln y)^2+y\ln y)\mathrm{d}x+(2x^2\ln y+x)\mathrm{d}y=0$ , change of variable $t=x\ln y$

I get that: $t' = \mathrm{d}t/\mathrm{d}x = \ln y + \frac{x}{y}y'$

$$\begin{align} -(2x^2\ln y+x)y' &= xy +2xy(\ln y)^2 + y \ln y \\ -(2x^2\ln y /y + x/y)y' &= x + 2x(\ln y)^2 + \ln y \\ -(2tx/y)y' &= x + 2x(\ln y)^2 + \ln y + y' x/y \\ -(2tx/y)y' &= x + 2t \ln y + t' \\ -(2t)(x/y) y' &= x +2t \ln y + t' \\ -(2t) (t' - \ln y) &= x + 2t \ln y + t' \end{align}$$ But it didn't get to something good to be easily integrable. Note that this question is asked for First-order Ordinary Differential Equation. So it probably shouldn't get very complicated.

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You just stopped too soon: $$\begin{aligned} -(2t)(t'-\ln y)&=x+2t\ln y+t'\\ -2tt'+2t\ln y&=x+2t\ln y+t'\\ -2tt'&=x+t'. \end{aligned}$$

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