Solution verification: $(2x\ln y)\mathrm{d}x+(\frac{x^2}{y}+3y^2)\mathrm{d}y=0$

Question

I've come across this problem while solving an exercise on Exact Differential equations and it's as follows:

Solve the differential equation: $$(2x\ln y)\mathrm{d}x+\left(\dfrac{x^2}{y}+3y^2\right)\mathrm{d}y=0$$

So what I've noticed is that if we expand the right side bracket as such: $$(2x\ln y)\mathrm{d}x+ \dfrac{x^2}{y}\mathrm{d}y+3y^2\mathrm{d}y=0 $$ Is that I can express the first two terms of the above equation as $$\mathrm{d}\left(x^2 \ln y\right)$$ Form here, simple integration leads us to: $$x^2\ln y + y^3= \mathrm{C}$$ Where $\mathrm{C}$ is the constant of integration. The solution on the other hand is given: $$x^2+\ln y+y^3=\mathrm{C}$$ Is this a misprint or am I missing something somewhere?

Answer 1

The solution given by the book can't be right, since differentiating gives

$$x^2+\ln y+y^3=C\implies 2x\,\mathrm dx+\left(\frac1y+3y^2\right)\,\mathrm dy=0$$

Your solution is correct.

Answer 2

$$(2x\ln y)\mathrm{d}x+ \dfrac{x^2}{y}\mathrm{d}y+3y^2\mathrm{d}y=0$$ $$\ln y{d}x^2+{x^2}d \ln y+3y^2\mathrm{d}y=0$$ $$d x^2\ln y+3y^2dy=0$$ Integration gives: $$x^2\ln y +y^3=C$$ Your answer is perfectly correct.