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Solve and find a particular solution that satisfies $y(1)=1$.

What is the type of this differential equation?

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    Homogeneous equation. – 2017-02-10
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    Quick observation: $x+y=0$ satisfies this equation – 2017-02-10
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    this is an equation of D'Alambert type – 2017-02-10
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    With respect to your second question, I'm sure if you consulted your text and/or course notes, you'd find your answer. The solving part .... please muster up the guts to at least try solving it on your own, and if you can't, have the guts to share where you're stuck. – 2017-02-10

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$$(x^2+xy)y'=x\sqrt{x^2-y^2}+xy+y^2$$ $$y'=\dfrac{x\sqrt{x^2-y^2}+xy+y^2}{(x^2+xy)}$$ is an homogeneous differential equation. Let $u=\dfrac{y}{x}$ so $$u'x+u=\dfrac{\sqrt{1-u^2}+u+u^2}{(u+1)}$$ or $$\dfrac{u+1}{\sqrt{1-u^2}}du=\dfrac{dx}{x}$$ after integration (let $u=\sin t$) we have $$t-\cos t=\ln Cx$$ with $y(1)=1$ we get $C=\exp(\dfrac{\pi}{2})$.

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    You made a mistake in re-arranging for $f(u)du = \frac{1}{x}dx$ – 2017-02-10
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    @Chinny84 Thanks. – 2017-02-10