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$$y~ dy+(2+x^2-y^2)dx$$

I try to solve this equation by putting standard form but becomes more challenge . So your answer is helpful

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    Is it equal to $0$ on the right hand side? – 2017-01-11
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    @projectilemotion: Likely is - it was unreadable before the edit - so maybe some stuff is missing. – 2017-01-11
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    it is a Bernoulli equation – 2017-01-11
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    see also here http://tutorial.math.lamar.edu/classes/de/bernoulli.aspx – 2017-01-11

2 Answers 2

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$y\,dy+(2+x^2-y^2)\,dx=0$
$y\,dy=-(2+x^2-y^2)\,dx$
$\frac{y\,dy}{dx}=-2-x^2+y^2$
$yy'-y^2=-2-x^2$

Use the following substitution: $v=y^2$. This gives us $v'=2yy'$

Now you have $\frac{1}{2}v'-v=-2x-x^2.$ It should be easy to take it from there.

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Expanding on emka's answer: \begin{align*} y\,dy+\left(2+x^2-y^2\right)dx=0\\ y\,dy=-\left(2+x^2-y^2\right)dx\\ y\frac{dy}{dx}=-2-x^2+y^2\\ \frac{d}{dx}\left(y^2\right)-y^2=-2-x^2 \end{align*} Make a substitution $v=y^2$: \begin{align*} \frac{dv}{dx}-v=-2-x^2\\ \frac{d^3v}{dx^3}-\frac{d^2v}{dx^2}=-2\\ \frac{\frac{d^3v}{dx^3}}{\frac{d^2v}{dx^2}-2}=1\\ \int\frac{1}{v''-2}\,dv''=\int1\,dx\\ \ln\left(v''-2\right)=x+C_1 \end{align*} Let $\exp(C_1)=c_1$: \begin{align*} v''=c_1e^x+2\\ v'=c_1e^x+2x+c_2\\ v=c_1e^x+x^2+c_2x\\ y^2=c_1e^x+x^2+c_2x \end{align*} Plugging back into the equation $\frac{d}{dx}\left(y^2\right)-y^2=-2-x^2$: \begin{align*} c_1e^x+2x+c_2-c_1e^x-x^2-c_2x=2-x^2\\ c_2-x^2+\left(2-c_2\right)x=2-x^2 \end{align*} Ergo $c_2=2$, yielding the following solution: $$y^2=c_1e^x+x^2+2x$$