Do solutions of $\dot{x} = \frac{x}{t^2} + t$ exist satisfying $x(0) =0$

Question

Suppose we have the 1-dimensional ODE \begin{equation} \dot{x} = \frac{x}{t^2} + t \end{equation} Do there exist solution curves with initial condition $x(0)=0$? If you proceed in a standard way then you would get as solution formula \begin{equation} x= C {\rm e}^{-1/t} + {\rm e}^{-1/t} \int^{t}_0 \tau {\rm e}^{1/\tau} d \tau \end{equation} but then the expression in the integral explodes! This does not imply that solution curves with $x(0)=0$ do not exist. How to proceed with such existence results?

Answer 1

The problem is that your original problem is not a Cauchy problem

The usual setup for solving differential equations is that you are given a function $f: [a,b] \times \mathbb R \to \mathbb R$ and a point $(x_0, y_0)$ and you want to find a function $y:[a,b]\to\mathbb R$ which satisfies two conditions:

• $f(x,y(x)) = y'(x)$ for all values $x\in[a,b]$
• $y(x_0)=y_0$

In your case, you only defined the function $f$ on $(a,b]\times\mathbb R$, since $f(t, x)=\frac{x}{t^2}+t$ in your case. Your equation has many solutions. For example, taking $x(t)=0$ creates a function that satisfies $f(t, x(t))=\dot x(t)$ for every $t\in (0,1]$.

Answer 2

Set $x(t)=t^2 y(t)$, then the original equation becomes:

$$t^2 \dot y(t)+2t y(t)=y(t)+t\tag{1}$$

The $y(0)=0$ solution is:

$$y(t)=\frac{1}{2}+\frac{1}{2t}-\frac{e^{-1/t}}{2t^2}Ei(1/t)\tag{2}$$

When $t\to 0$, we have

$$y(t)\approx -t - 3 t^2 - 12 t^3 - 60 t^4...\tag{3}$$