Let y=y(x), y0, solve (1+x^2),dy=y(x-y),dx.

Question

Let $y=y(x)$, $y>0$, solve $(1+x^2)\,dy=y(x-y)\,dx$. If $y(0)=1$ and $y(2\sqrt2)=\beta$, which relation holds:

Accepted Answer

Dividing by $y^2$ and substituting $t=1/y$ gives a linear ODE:
$$\frac{dt}{dx}+\frac{x}{1+x^2}t = \frac{1}{1+x^2}.$$

I.F.$=\sqrt{1+x^2}$. Solving: $t\sqrt{1+x^2} = \ln\left(x+\sqrt{1+x^2}\right)+c$.

Using $y(0)=1 \Rightarrow t(0)=1 \Rightarrow c=1$. At $x=2\sqrt2$, $\sqrt{1+x^2}=3$, giving
$$\frac{3}{\beta} = \ln\big(2\sqrt2+3\big)+1 \implies e^{3/\beta-1}=e(3+2\sqrt2)\ \Rightarrow e^{3/\beta}=e(3+2\sqrt2).$$

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