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I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$

Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$

It is not exact since partial derivatives are not equal.

Any help would be appreciated.

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    They are equal though. – 2017-02-13
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    @AhmedS.Attaalla, there is a minus right before the 2nd bracket. – 2017-02-13

3 Answers 3

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To put it into the form you requested: $$ -(1+x^2) \,dy + (2xy + x^2 + x^4) \,dx = 0 \implies \frac{dy}{dx} - \frac{2xy + x^2 + x^4}{1+x^2} = 0 \\ \implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = \frac{x^2 + x^4}{1+x^2}\\ \implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = x^2 $$

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    Beat me by a minute, and with derivation too. ;) +1 – 2017-02-13
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    Thank you! It was pretty easy, donno how I couldn't figure it out by myself – 2017-02-13
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Hint

We can write your equation as $y' - \frac{2xy}{x^2+1} = x^2$

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$$(2xy+x^2+x^4)\,dx−(1+x^2)\,dy=0\\ (1+x^2) \frac {dy}{dx} = (2xy+x^2+x^4)\\ y' - \frac {2x}{1+x^2} y = x^2$$

Integrating factor

$$e^{\int \frac {-2x}{1+x^2}} = e^{-\ln(1+x^2)} = \frac {1} {(1+x^2)}$$

$$\frac {1}{1+x^2}y = \int \frac {x^2}{1+x^2}\\ \frac {1}{1+x^2}y = x - \arctan x + C\\ y = x^3 + x - (1+x^2)\arctan x + C(1+x^2)$$