(e^x)y' = sin(e^-x) + (e^x)(cos(e^-x)) - e^(-x) - (e^x)(y)?

Question

I used the integrating factor method and got

d/dx (ye^x) = sin(e^-x) + (e^x)(cos(e^-x)) - e^(-x)

Now comes the problem, how do you integrate sin(e^-x) ?

Answer 1

♠ Let’s brush it first: t=exp(-x) and x=-ln(t) and dx=-dt/t;
thus your equation looks:
dy/t = -(sint +cost/t –t –y/t)dt/t; t·y’= -t·sint –cost +t·t +y;
♣ t·y’-y= -t·sint -cost +tt; (t·y’-y)/tt = 1-sint/t –cost/tt; (y/t)’ =1-sint/t –cost/tt;
♦ thus y/t =t -s1-s2, where s1=∫dt·sint/t, s2=∫dt·cost/tt;
♥ now run your fingers and do find s1 with s2. When done the result will be near. If embarrassed again just click my name and write me, you are welcome!

Answer 2

If you make the change z= e^-x dz = -e^-x dx so dx=-dz/z You come to
Integral -sin(z)/(z)*dz which can not be written as a combination of so called elementary functions

Answer 3

isn't it just cos(e^-x) * e^-x?