Solving Non-Exact Equations (2x-4y+6)dx+(x+y-3)dy=0

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(2x-4y+6)dx+(x+y-3)dy=0
i can't make integration cofficient
because

(P_y-Q_x)/P is not a function of x and so is the other one

??

for both (P_y-Q_x)/P and (P_y-Q_x)/Q i don't get a function of one variablr i can't make it exact

how to solve it
 
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Try a substitution of y=Vx
 
if i substite by y=vx
what is dy
?
 
Are we allowed to assume that you have taken calculus?
Use the product rule to differentiate y= vx
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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