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

Question

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

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

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

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Answer 1 · 2018-02-18T08:30:51

$Given that, e^{- x} \frac{d y}{d x} - e^{- x} y = e^{- x} cos x$ $"Given that, "e^-xdy/dx-e^-xy=e^-xcosx$ .

Note that, by the Chain Rule, $\frac{d}{d x} (e^{- x}) = e^{- x} \cdot \frac{d}{d x} (- x)$ $d/dx(e^-x)=e^-x*d/dx(-x)$ ,

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

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,

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