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I am trying to solve non exact differential equation but I am unable to find the integration factor.

I solve the following question $$-3y\frac{dy}{dx}+2x=0 $$

and I get the integrationg factor $x^{-5/2}$ and I use the proper way to find the integrating factor, but in our text book the integrating facot for above differential equation is finded and that is $y/x^4$.

Now I am completely confused that why please help me out with this.

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    $\begingroup$ Why do you need an integrating factor, you can directly integrate to $-\frac32y^2+x^2=C$. $\endgroup$ Dec 6 '20 at 8:43
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$$-3y\:dy+2x\:dx=0$$ is an EXACT differential equation.

In other words the integrating factor is $1$. $$d\left(-\frac32 y^2+x^2\right)=0$$ $$-\frac32 y^2+x^2=\text{constant}$$

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$$-3y\frac{dy}{dx}+2x=0$$ $$-\dfrac 32 2yy'+2x=0$$ Note that $$((y^2)'=2yy'$$ So that you have: $$-\dfrac 32 (y^2)'+2x=0$$ Integrate: $$-\dfrac 32 y^2+x^2=c$$

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