What role does dummy variable play in this example?

Question

It's an IVP in Separation Equations : $$\frac {dy}{dx}=e^{-x^2}, y(3)=5$$ Here is the solution on text: $$\int^x_3\frac {dy}{dt}dt=\int_3^xe^{-t^2}dt$$ It comes out $$y(x)=y(3)+\int_3^xe^{-t^2}dt$$ And here is my answer with steps $$\int^x_3dy=\int^x_3e^{-x^2}dx$$ $$y]^x_3=\int^x_3e^{-x^2}dx$$ $$y(x)-y(3)=\int^x_3e^{-x^2}dx$$ And they were in my opinion the same. So why bother using dummy variable? Did I ignore any concept?

Answer 1

The reason is that we must not mixed up the limits of an integral and the differential coefficient which denotes by which the integration will be evaluated.

For example, $$\int_{1}^21d1$$ is meaning less no one treats $1$ as a variable.

Another is $$\int_{0}^xx^2dx$$ is meaning less as the limit is $x$ and we are integrating w.r.t. $x$. Confusion may arise since after evaluation we get $$\left[\frac{t^3}{3}\right]_{x=0}^{x=x}$$ see that $x=x$ term?? Very confusing as it gives $0=0$. That's why Dummy variable is used i.e. $$\int_{0}^xt^2dt=\left[\frac{x^3}{3}\right]_{t=0}^{t=x}$$ see? No classical confusion arises for this notation.