How does $e^(-x)(dy/dx)-e^(-x)y=e^(-x)cosx$ simplify to $d/dx(ye^-x)=e^-xcosx$ ?

Question

In a differential equation example in my book,
$e^(-x)dy/dx-e^-xy=e^(-x)cosx$
$d/dx(ye^-x)=e^-xcosx$

How do you get the second step from the first, I don't understand?

How does $e^(-x)(dy/dx)-e^(-x)y=e^(-x)cosx$ simplify to $d/dx(ye^-x)=e^-xcosx$ ?

1204 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-18T08:30:51

$"Given that, "e^-xdy/dx-e^-xy=e^-xcosx$ .

Note that, by the Chain Rule, $d/dx(e^-x)=e^-x*d/dx(-x)$ ,

$:. d/dx(e^-x)=-e^-x$ .

Using this, we replace the second term $e^-x$ on the left member of

the eqn., and, get,

$e^-x*d/dx(y)+y*d/dx(e^-x)=e^-xcosx$ .

Observe that, by the Product Rule, the left member is,

$d/dx(y*e^-x)$ .

Hence, the eqn. becomes,

$d/dx(y*e^-x)=e^-x*cosx$ .