Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 3

Recall that an ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows (dy)/(dx) = f(x,y) .
In the given problem: x^2+5y(dy)/(dx)=0 , we apply variable separable differential equation in a form of f(y) dy = f(x) dx .
Move the x^2 to the other side: 5y(dy)/(dx)=-x^2
Transfer the (dx) to the other side by cross-multiplication: 5y dy=-x^2 dx
Apply direct integration: int5y dy=int-x^2 dx
Apply the basic integration property: int c*f(x) dx= c int f(x) dx .
5int y dy=(-1) intx^2 dx
Apply Power Rule of integration: int x^ndx= x^(n+1)/(n+1) .
5*y^(1+1)/(1+1)=(-1) * x^(2+1)/(2+1)+C
(5y^2)/2=-x^3/3+C
Multiply both side by 2/5, we get:
(2/5)(5y^2)/2=(2/5)(-x^3/3+C)
Note: (2/5)*C = C since C is an arbitrary constant.
y^2=(-2x^3)/15+C
y=+-sqrt(-(2x^3)/15+C)