# xy + y' = 100x Solve the differential equation 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)