MHB Solution of differential equation

[evinda]

Hello! (Wave)

I am looking at the following exercise:

We consider the differential equation $x^2y'+2xy+1=0$, where $0<x< +\infty$.

• Show that each solutions goes to $0$ while $x \to +\infty$.
• Find the solution $\phi$ of the above differential equation so that $\phi(2)=2 \phi(1)$.

That's what I have tried:

The differential equation $x^2y'+2xy+1=0$ can be written as $(x>0)$:
$$y'+\frac{2y}{x}=\frac{1}{x^2}$$
We fix a random $x_0 \in (0,+\infty)$, let $x_0=1$.

Then $A(x):=\int_1^x \frac{2}{t}dt=2 \ln x$Therefore, if $\phi$ is a solution of the differential equation then:

$$e^{A(x)} \phi'(x)+\frac{2}{x}e^{A(x)}\phi(x)=\frac{e^{A(x)}}{x^2}$$
$$\Rightarrow x^2 \phi'(x)+2x \phi(x)=1\\ \Rightarrow (x^2 \phi(x))'=1, x \in (0,+\infty)\\ (t^2 \phi(t))'=1, t \in (0,+\infty)$$

We integrate the above relation from $1$ to $x$ (for each $x \in (0,+\infty)$) and we have:

$$x^2 \phi(x)-\phi(1)=\int_1^x 1 dt=x-1$$

and so $\phi(x)=\frac{1}{x}+\frac{1}{x^2}(\phi(1)-1)$

$\frac{1}{x^2}$ is a solution of the homogeneous differential equation $y'+\frac{2y}{x}=0$ and $\frac{1}{x}$ is a solution of $y'+\frac{2y}{x}=\frac{1}{x^2}$

So, each solution $\phi$ of the differential equation $y'+\frac{2y}{x}=\frac{1}{x^2}$ is of the form:

$$\phi(x)=\frac{1}{x}+c \frac{1}{x^2}$$

$$\lim_{x \to +\infty} \phi(x)=0$$

$$ \phi(1)=1+c \\ \phi(2)=\frac{1}{2}+c \frac{1}{4}\\ \phi(2)=2 \phi(1) \Rightarrow c=-\frac{6}{7}$$Is the way I found the general solution right or shouldn't I integrate $(t^2 \phi(t))'=1, t \in (0,+\infty)$ from $1$ to $x$? (Thinking)

 

[MarkFL]

I would write the given ODE as:

$$\frac{d}{dx}\left(x^2y\right)=-1$$

Integrating with respect to $x$, we obtain:

$$x^2y=C-x$$

Hence:

$$y(x)=\frac{C-x}{x^2}$$

We can see that:

$$\lim_{x\to\infty}y=0$$

Now, using the initial conditions, we find:

$$\frac{C-2}{4}=2(C-1)\implies C=\frac{6}{7}$$

Hence:

$$\phi(x)=\frac{6-7x}{7x^2}$$