Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 10

The general solution of a differential equation in a form of f(y) y'=f(x) can
be evaluated using direct integration.
We can denote y' as (dy)/(dx) then,
f(y) y'=f(x)
f(y) (dy)/(dx)=f(x)
Rearrange into : f(y) (dy)=f(x) dx
To be able to apply direct integration : intf(y) (dy)=int f(x) dx.
Applying this to the given problem: yy'=-8cos(pix) , we get:
y(dy)/(dx)=-8cos(pix)
y(dy)=-8cos(pix)dx
int y(dy)=int-8cos(pix)dx
For the integration on the left side, we apply Power Rule integration: int u^n du= u^(n+1)/(n+1) on int y dy .
int y dy = y^(1+1)/(1+1)
= y^2/2
For the integration on the right side, we apply the basic integration property: int c*f(x)dx= c int f(x) dx and basic integration formula for cosine function: int cos(u) du = sin(u) +C
int -8 cos(pix) dx= -8 int cos(pix) dx
Let u = pix then du = pi dx or (du)/pi=dx.
Then the integral becomes:
-8 int cos(pix) dx=-8 int cos(u) *(du)/pi
=-8/pi int cos(u) du
=-8/pi*sin(u) +C
Plug-in u=pix in -8/pi*sin(u) +C , we get:
-8 int cos(pix) dx=-8/pi*sin(pix) +C

Combing the results, we get the general solution for differential equation (yy'=-8cos(pix)) as:
y^2/2=-8/pi*sin(pix) +C
2* [y^2/2] = 2*[-8/pi*sin(pix)]+C
y^2 =-16/pi*sin(pix)+C
The general solution: y ^2=-16/pisin(pix)+C can be expressed as:
y = +-sqrt(-16/pisin(pix)+C) .