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$$ \frac{dy}{dx}=\frac{x(x^2+y^2+1)}{2y(x^2+1)} $$ Spent about 2 hours but still cant find any proper way. Any suggestions? Thank you.

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With $z:=y^2$, this is a linear equation

$$(x^2+1)z'-xz=x(x^2+1).$$

The homogeneous part is separable,

$$\frac{z'}{z}=\frac{x}{x^2+1}$$ giving $$z=a\sqrt{x^2+1}.$$

Now by variation of the constant, $z=a(x)\sqrt{x^2+1}$, we get

$$(x^2+1)^{3/2}a'(x)=x(x^2+1)$$ and $$a(x)=\sqrt{x^2+1}+b.$$

Finally,

$$y^2=x^2+1+b\sqrt{x^2+1}.$$

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Hint:

Multiply both sides with $2y$, then substitute $y(x)^2=f(x)$

$\frac{\partial (y^2)}{\partial x}$ becomes $2\frac{\partial y}{\partial x} \cdot y$

Maybe you can finish from there. Solve in terms of $f$ and then substitute back.

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$$\frac{dy}{dx}=\frac{x(x^2+y^2+1)}{2y(x^1+1)} \implies \frac{2y}{x}\frac{dy}{dx}=\frac{x^2+y^2+1}{x^2+1}.$$ Ket $x^2+1=X$, $y^2=Y$, then the ODE becomes $$2\frac{dY}{dX}=\frac{X+Y}{X} \implies \frac{dY}{dX}-\frac{Y}{2X}=\frac{1}{2}.$$ For this the integrating Factor is $I=\frac{1}{\sqrt{X}}$, the solution is $$Y=\sqrt{X} \int \frac{dX}{2\sqrt{X}}+C\sqrt{X} \implies Y=X+C \sqrt{X}.$$ So finally the so;ution is $$y^2=x^2+1+C\sqrt{x^2+1}$$

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$$\frac{dy}{dx}=\frac{x(x^2+y^2+1)}{2y(x^2+1)}$$ $$\frac{dy}{x(x^2+y^2+1)}=\frac{dx}{2y(x^2+1)}$$ $$\frac{dy^2}{(x^2+y^2+1)}=\frac{dx^2}{2(x^2+1)}$$ $$\frac{du}{(v+u)}=\frac{dv}{2v}$$ Where $u=y^2$ an $v=x^2+1$ I think it's more simple now to solve. $$u'=\frac 12 \left(1 +\frac u v \right)$$ it's a first order linear DE. Can you take it from here ?

You can also do this: $$\frac{du}{(v+u)}=\frac{dv}{2v}$$ $$\frac{d(u-v)}{(u-v)}=\frac{dv}{2v}$$ Integrate: $$\ln (u-v)=\frac 12 \ln(v) +C$$ $$\boxed {\implies u(v)=v+C \sqrt v}$$ Substitute back $u,v$

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