Calculus of a Single Variable, Chapter 6, 6.1, Section 6.1, Problem 52

The given problem: (dy)/(dx) = 5e^(-x/2) is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: N(y) dy= M(x)dx
Cross-multiply dx to the other side, we get:
dy= 5e^(-x/2)dx
In this form, we may now proceed to direct integration on both sides:
int dy= int 5e^(-x/2)dx
For the left side, we apply basic integration property: int (dy)=y .
For the right side, we may apply u-substitution by letting: u = -x/2 then du =-1/2 dx or -2du= dx .
Plug-in the values: -x/2=u and dx=-2du , we get:
int 5e^(-x/2)dx=int 5e^(u)* (-2 du)
=int -10e^(u)du
Apply the basic integration property: int c*f(x)dx= c int f(x) dx .
int -10e^(u) du=(-10) int e^(u) du
Apply basic integration formula for exponential function:
(-10)int e^(u) du= -10e^(u)+C
Plug-in u=-x/2 on -10e^(u)+C , we get:
int 5e^(-x/2) dx=-10e^(-x/2)+C
Combining the results from both sides, we get the general solution of differential equation as:
y=-10e^(-x/2)+C