Show that only one solution of xy'-(2x^2+1)y=x^2 tends to a…
Show that only one solution of $xy'-(2x^2+1)y=x^2$ tends to a finite value as $x\to\infty$, and find that solution.
Explanation of Method
Divide by $x$ to get standard linear form and integrate with the Gaussian-type factor $e^{-x^2}/x$; the resulting general solution involves $\int e^{-t^2}dt$, and only one choice of the constant keeps it bounded at infinity.
Step 1: Solve the linear equation
$y'-\left(2x+\dfrac1x\right)y=x$. The integrating factor is $\mu(x)=e^{-x^2}/x$ (for $x>0$), giving $\dfrac{d}{dx}\!\left(\dfrac{e^{-x^2}}{x}y\right)=e^{-x^2}$.
Step 2: Integrate using the limits $x$ to $\infty$
Choosing the antiderivative $-\!\int_x^\infty e^{-t^2}dt$ (which vanishes appropriately) gives $\dfrac{e^{-x^2}}{x}y=-\!\displaystyle\int_x^\infty e^{-t^2}\,dt+K$, so
\[y(x)=-xe^{x^2}\int_x^\infty e^{-t^2}\,dt+Kxe^{x^2}.\]
Step 3: Force a finite limit as $x o\infty$
Since $\displaystyle\int_x^\infty e^{-t^2}dt\sim\frac{e^{-x^2}}{2x}$ for large $x$, the first term tends to $-\tfrac12$, while $Kxe^{x^2}\to\infty$ unless $K=0$. So $K=0$ is forced.
Result
Final answer:
\[\boxed{\,y(x)=-xe^{x^2}\int_x^\infty e^{-t^2}\,dt,\qquad \lim_{x\to\infty}y(x)=-\frac12.}\]
Every other value of the constant makes $y(x)\to\pm\infty$.