Calculus of a Single Variable, Chapter 6, 6.2, Section 6.2, Problem 10

The problem: xy+y'=100x is as first order differential equation that we can evaluate by applying variable separable differential equation:
N(y)y'=M(x)
N(y)(dy)/(dx)=M(x)
N(y) dy=M(x) dx
Apply direct integration: intN(y) dy= int M(x) dx to solve for the
general solution of a differential equation.
Applying variable separable differential equation, we get:
xy+y'=100x
y' =100x-xy
y'=x(100-y)
(y')/(100-y)= x
Let y' =(dy)/(dx) :
((dy)/(dx))/(100-y)= x
(dy)/(100-y)= x dx
Apply direct integration on both sides:
int(dy)/(100-y)= int x dx
For the left side, we consider u-substitution by letting:
u= 100-y then du = -dy or -du=dy.
The integral becomes:
int(dy)/(100-y)=int(-du)/(u)
Applying basic integration formula for logarithm:
int(-du)/(u)= -ln|u|
Plug-in u = 100-y on "-ln|u| " , we get:
int(dy)/(100-y)=-ln|100-y|

For the right side, we apply the Power Rule of integration: int x^n dx = x^(n+1)/(n+1)+C

int x* dx= x^(1+1)/(1+1)+C

= x^2/2+C

Combing the results from both sides, we get the general solution of the differential equation as:
-ln|100-y|= x^2/2+C
or
y =100- e^(-x^2/2-C)
y = 100-Ce^(-x^2/2)