About Differential Equation. How can I prove $y(x)\neq 0$?

Question

This is a familiar differential equation.

\begin{cases} \dfrac{dy}{dx}=y \\ y(0)=1 \\ \end{cases}

When we solve this equation, we will do the following operation.

From $\dfrac{dy}{dx}=y,$ we get $\dfrac{1}{y} \dfrac{dy}{dx}=1.$

But when we do this, we have to mention that $y(x)\neq 0$ for all $x$.

I wonder how I can prove $y(x)\neq 0$ for all $x$.

This is my attempt.

Suppose there exists $x_1$ such that $y(x_1)=0$.

Because $y(0)=1,$ $x_1\neq 0$.

And we can say that $\dfrac{dy}{dx}(x_1)=y(x_1)=0.$

I couldn't proceed from this.

I think that I have to use "the uniqueness of solution of differential equations". But I don't know where I can use this.

I would like you to give me some ideas.

Answer 1

If you allow knowledge of the exponential function, $e^x$ then you can use an "integrating factor". You have the equation $\displaystyle \frac{dy}{dx} = y$. Write it as $$ \frac{dy}{dx}-y = 0 $$ and multiply by $e^{-x}$ to get, $$ e^{-x} \frac{dy}{dx} - e^{-x} y = 0. $$ Apply the product rule to see that the left side is now a differential, and, $$ \frac{d}{dx}\Big( e^{-x}y \Big) = 0. $$ You only need now to know that the derivative of a function is zero if and only if the function is constant, and the constant is set from your initial condition. Thus $y = e^x$ which we know to be non-zero.

Alternatively, if you want to avoid using knowledge of the exponential function but you do know that such differential equations have a unique solution given any initial condition $y(x_0)$ at some $x=x_0$ you can argue that if $y(x)$ is a solution that ever equals zero at a point, $x_0$ say, it must be zero for all $x$ because the zero function is the unique solution satisfying the given condition. Therefore a solution cannot also satisfy $y(0)=1$.

Answer 2

This is a slightly incomplete answer, but it should be helpful.

As $y$ is differentiable, it is also continuous. Since $y(0) = 1 > 0$, the only way for $y$ to be zero for some $x_0 > 0$ is if $\frac{dy}{dx}$ is negative at some point. However $\frac{dy}{dx} = y$ so for small $x > 0$ (small enough to guarantee $y$ is still positive (continuity is important here)), $\frac{dy}{dx}$ will be positive on $[0,x]$ (since $y'$ and $y$ have the same sign) and thus $y(x) > y(0)$ by the mean value theorem. But then there is another interval on which $y$ is positive and so on. It's easy to see that $y$ will never be $0$.

What about $x < 0$? Again we know that an interval $[x, 0]$ exists such that $y$ is positive here since $y(0) = 1$. However since $y'$ is positive on $[x,0]$, we know by mean value theorem that $0 < y(x) < y(0)$. Consider another interval with upper bound $x$ instead and another and so on.

There is a slight caveat here: this does not prove that the solution exists for all $x\in\mathbb{R}$, but rather where it does exist, it must be positive. You could imagine that each successive interval in both iterative processes gets smaller and smaller so that you cannot exhaust the entire real line. I am sure a modest modification of what I've written can be turned into a full proof but I can't see it immediately.

Answer 3

as you said, if: $$\frac{dy}{dx}=y\Rightarrow \int\frac1y\frac{dy}{dx}dx=\int dx\Rightarrow \ln|y|=x+C_1$$ $$y=C_2e^x$$ and so for $y=0$ at any point $x$ would require $C_2=0$ but $C_2=e^{C_1}$ and so: $C_1\to-\infty$. All of this gives us that if $|x|<\infty,y\ne0$ or it will contradict $y(0)=1$