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In class the teacher wrote the following that I am having trouble accepting.

We have a differential equation $$\frac{1}{x(s)}\frac{dx(s)}{ds}=f(y)\frac{dy(s)}{ds},$$ which gets rewritten as $$\frac{d\text{log}(x(s))}{ds}=f(y)\frac{dy(s)}{ds}.$$

They then proceed to multiply across by $ds$ and integrate the left hand side to get $$\text{log}(x(s))=\int f(y)\frac{dy(s)}{ds} ds=\int f(y)\,dy(s).$$

Can you just do this like this? It seems weird to me that we can just go from something that's a derivative to something that's not by multiplying across by a form.

furious.neutrino
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1 Answers1

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Yes. Anti-derivatives and derives cancel out. You can verify that by doing it in reverse: $$\ g(x(s))=\int f(y)\,dy(s)$$ $$\ \frac{d}{ds} g(x(s))= \frac{d}{ds} \int f(y)\,dy(s)$$ By chain rule, $$\ \frac{dx(s)}{ds} \cdot \frac{d}{dx(s)} g(x(s))= \frac{dy(s)}{ds} \cdot \frac{d}{dy(s)} \int f(y)\,dy(s) $$ $$\ g’(x(s)) \frac{dx(s)}{ds} = f(y) \frac{dy(s)}{ds} $$ Take $g(x)=log(x)$, it yields exactly your work in reverse. The way they “multiply by dx” is actually just a handy way to solve the differential equations.

Mr. W
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