Given differential equation is
(ex+e−x)dy−(ex−e−x)dx=0
(ex+e−x)dy=(ex−e−x)dx
dy=ex−e−xex+e−xdx
Integrating both sides
We get,
∫ dy=∫ex−e−xex+e−xdx (i)
Let u=ex+e−x
Differentiate u w.r.t u
we get,
dudx=(ex−e−x)
dx=duex−e−x (a)
Putting value of (a) in (i) then we get,
∫ dy=∫ex−e−xuduex−e−x
∫dy=∫duu
y=log|u|+c
Now, putting u=ex−e−x then we get,
y=log|ex−e−x|+c
Final Answer:
Hence, the required general solution is
y=log|ex−e−x|+c