Different equation with bounded condition

Question

$$y' = 2x(π+y)$$ $y$ being bounded when $x\to + ∞$

There is a singular solution $~y = -π~$ and it was the answer.

But general solution $y = ce^{x^2} - π$ ;$(|c| = e^{c})$ and $~c~$ must not be $~0~$.

$y - π = 0 = ce^{x^2}$ then $~c = 0~$? why $~c~$ can be $~0~$ in this case?

And how do you know $~y = -π~$ satisfy the bounded condition?

Answer 1

$$y'-2xy=2x\pi$$ $$IF=e^{-x^2}$$

$$\int d(ye^{-x^2})=2\pi\int xe^{-x^2}dx+C $$ $$ye^{-x^2}=-2\pi(\frac{e^{-x^2}}{2})+C \tag{1}$$ as $x\rightarrow \infty$ $C\rightarrow0$ $$y=-\pi$$

Hope it solved it for you?

Answer 2

Put $z=y+\pi $. then $$\frac {z'}{z}=2x $$ and

$$z=\lambda e^{x^2} $$

$$y(x)=\lambda e^{x^2}-\pi $$

if $\lambda\neq 0$ then

$$\lim_{x\to+\infty}y (x)=\infty $$ but $y $ is bounded, thus $\lambda=0$ and $$y (x)=-\pi $$ is the only bounded solution.