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Solve using integrating factor: $$ y dx - (y^2+x^2+x) dy=0$$

My attempt first get the D.E into the form

$\frac{dy}{dx} +p(x)y(x)=q(x)$ so

So my integrating factor is $e^{\int p(x) dx}$

$$ ydx - (y^2+x^2+x) dy=0$$

$$ -ydx + (y^2+x^2+x)dy=0$$

$$ \frac{-y}{y^2+x^2+x} + \frac{dy}{dx}=0$$

$$ \frac{dy}{dx} - \frac{1}{y^2+x^2+x} \cdot y =0 $$

However how can I have an integrating factor of

$e^{\int - \frac{1}{y^2+x^2+x} dx}$ ?

Where am I going wrong?

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    p must be a function of x, not y – 2017-01-02
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    @vvnitram yes I understand that but how I do get my d.e into that form , it seems like I cannot do that – 2017-01-02
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    i have got an ugly solution – 2017-01-02

1 Answers 1

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You can follow this method via use of exact differentials, if you like:

$$ y dx - (y^2+x^2+x) dy=0$$ $$ y dx - x dy =(y^2+x^2) dy$$ $$ \frac{y dx - x dy}{(y^2+x^2)} = dy$$ $$d\left(\arctan\frac{x}{y}\right)=dy$$ $$\boxed{y=\left(\arctan\frac{x}{y}\right)+c}$$

This is the required solution.

Hope this helps you.

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    Thank you. My book however says the integrating factor should be $\frac{1}{x^2+y^2}$ how is this possible? – 2017-01-02