How does e^(-x)(dy/dx)-e^(-x)y=e^(-x)cosxex(dydx)exy=excosx simplify to d/dx(ye^-x)=e^-xcosxddx(yex)=excosx?

In a differential equation example in my book,
e^(-x)dy/dx-e^-xy=e^(-x)cosxexdydxexy=excosx
d/dx(ye^-x)=e^-xcosxddx(yex)=excosx

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)cosxex(dydx)exy=excosx simplify to d/dx(ye^-x)=e^-xcosxddx(yex)=excosx?

1 Answer
Feb 18, 2018

"Given that, "e^-xdy/dx-e^-xy=e^-xcosxGiven that, exdydxexy=excosx.

Note that, by the Chain Rule, d/dx(e^-x)=e^-x*d/dx(-x)ddx(ex)=exddx(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.