About Differential Equation. How can I prove $y(x)\neq 0$? 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 &quot;the uniqueness of solution of differential equations&quot;. But I don't know where I can use this.
I would like you to give me some ideas.
If you allow knowledge of the exponential function, $e^x$ then you can use an &quot;integrating factor&quot;.  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 $y(0)=1$.
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 $|x|&lt;\infty,y\ne0$ or it will contradict $y(0)=1$