Show that every solution ϕ of $x^2y'+2xy = 1$ on $0

Question

Please help me with the following proof:

Consider the equation $x^2y'+2xy = 1$ on $0

Show that every solution tends to zero as $x→∞$. *General solution: $ϕ(x) = \frac{1}{x} + \frac{c}{x^2}$.

Answer 1

Rewrite the given differential equation as $(x^2y)'=1$. This implies $x^2y=x+c$ $\ (x>0)$ for some constant $c$, hence $y(x)={1\over x}+{c\over x^2}$, as indicated in the question. It follows that $\lim_{x\to\infty}y(x)=0$, whatever $c$.

Answer 2

It is sufficient to show that for and real $c$, any positive real $\epsilon$ there is another positive real number $b$ such that for every $x>b$ the inequality below holds $$ \frac{1}{x} + \frac{c}{x^2}<\epsilon$$

Answer 3

To solve this first-order linear ordinary differential equation, substitute:

$$2x=\frac{\text{d}}{\text{d}x}\left(x^2\right)\tag1$$

So, we get that:

$$x^2\cdot\text{y}'\left(x\right)+2x\cdot\text{y}\left(x\right)=1\space\Longleftrightarrow\space x^2\cdot\text{y}'\left(x\right)+\frac{\text{d}}{\text{d}x}\left(x^2\right)\cdot\text{y}\left(x\right)=1\tag2$$

Now, apply the reverse product rule:

$$\int\frac{\text{d}}{\text{d}x}\left(x^2\cdot\text{y}\left(x\right)\right)\space\text{d}x=\int1\space\text{d}x=x^2\cdot\text{y}\left(x\right)=x+\text{C}\space\Longleftrightarrow\space\text{y}\left(x\right)=\frac{x+\text{C}}{x^2}\tag3$$

Now, we can apply l'Hôpital's rule:

$$\lim_{x\to\infty}\text{y}\left(x\right)=\lim_{x\to\infty}\frac{x+\text{C}}{x^2}=\lim_{x\to\infty}\frac{1+0}{2\cdot x}=\frac{1}{2}\cdot\lim_{x\to\infty}\frac{1}{x}=\frac{1}{2}\cdot0=0\tag4$$