Question

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

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`?

Answer 1

`"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`.