We have:
\(\dfrac{\delta y}{\delta x}=f(x)g(x)\)
We want that in terms of \(y\).
We know from the product rule of differentiation:
\(y=a(x)b(x)\)
Means that:
\(\dfrac{\delta y}{\delta x}=a'(x)b(x)+a(x)b'(x)\)
So let’s relabel \(f(x)\) as \(h'(x)\)
\(\delta\)
\(\dfrac{\delta y}{\delta x}=h'(x)g(x)\)
\(\dfrac{\delta y}{\delta x}+h(x)g'(x)=h'(x)g(x)+h(x)g'(x)\)
\(y+\int h(x)g'(x)=\int h'(x)g(x)+h(x)g'(x)\)
\(y+\int h(x)g'(x)=h(x)g(x)\)
\(y=h(x)g(x)-\int h(x)g'(x)\)
For example:
\(\dfrac{\delta y}{\delta x}=x.\cos(x)\)
\(f(x)=\cos(x)\)
\(g(x)=x\)
\(h(x)=\sin(x)\)
\(g'(x)=1\)
So:
\(y=x\int \cos(x) dx-\int \sin(x)dx\)
\(y=x\sin(x)-\cos(x)+c\)