Answer:
y(x) = -1/(x^2 + c_1 x)
Explanation:
Solve Bernoulli's equation ( dy(x))/( dx) + y(x)/x = x y(x)^2:
Divide both sides by -y(x)^2:
-(( dy(x))/( dx))/y(x)^2 - 1/(x y(x)) = -x
Let v(x) = 1/y(x), which gives ( dv(x))/( dx) = -(( dy(x))/( dx))/y(x)^2:
( dv(x))/( dx) - v(x)/x = -x
Let μ(x) = e^( integral-1/x dx) = 1/x.
Multiply both sides by μ(x):
(( dv(x))/( dx))/x - v(x)/x^2 = -1
Substitute -1/x^2 = d/( dx)(1/x):
(( dv(x))/( dx))/x + d/( dx)(1/x) v(x) = -1
Apply the reverse product rule f ( dg)/( dx) + g ( df)/( dx) = d/( dx)(f g) to the left-hand side:
d/( dx)(v(x)/x) = -1
Integrate both sides with respect to x:
integral d/( dx)(v(x)/x) dx = integral-1 dx
Evaluate the integrals:
v(x)/x = -x + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(x) = 1/x:
v(x) = x (-x + c_1)
Solve for y(x):
y(x) = 1/v(x) = -1/(x^2 - c_1 x)
Simplify the arbitrary constants:
Answer: y(x) = -1/(x^2 + c_1 x)