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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?

savemycalculus
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  • The issue is that as it is you've used $x$ as your variable for integration and in the bounds, so it's both free and bound in that context which really just doesn't make any sense from a notational standpoint. We use the dummy variable for the integration because it makes it a lot clearer which $x$'s are for our integration and which refer to the more global way you use $x$ here. – Stephen Donovan Jun 29 '21 at 02:50
  • @StephenDonovan but on the LHS. Why using (dy/dt)/dt? that is so unnecessary. – savemycalculus Jun 29 '21 at 03:37

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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.

Sujit Bhattacharyya
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  • Thanks, I knew that and I was not very precise up there, the real problem is on the LHS where it wrote (dy/dt)dt. I can't understand the purpose of this work. – savemycalculus Jun 29 '21 at 03:29