Prove that equation is a Bernoulli equation and solve:

Question

dy/dx + y/x = (x³)(y²)

Answer 1

Solve Bernoulli's equation (dy(x))/(dx) + y(x)/x = x^3 y(x)^2:

Divide both sides by -y(x)^2:

-((dy(x))/(dx))/y(x)^2 - 1/(x y(x)) = -x^3

Let v(x) = 1/y(x), which gives (dv(x))/(dx) = -((dy(x))/(dx))/y(x)^2:

(dv(x))/(dx) - v(x)/x = -x^3

Let μ(x) = e^( integral-1/x dx) = 1/x.

Multiply both sides by μ(x):

((dv(x))/(dx))/x - v(x)/x^2 = -x^2

Substitute -1/x^2 = d/(dx) (1/x):

((dv(x))/(dx))/x + d/(dx) (1/x) v(x) = -x^2

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) = -x^2

Integrate both sides with respect to x:

integral d/(dx) (v(x)/x) dx = integral-x^2 dx

Evaluate the integrals:

v(x)/x = -x^3/3 + c_1, where c_1 is an arbitrary constant.

Divide both sides by μ(x) = 1/x:

v(x) = x (-x^3/3 + c_1)

Solve for y(x):

y(x) = 1/v(x) = -3/(x^4 - 3 c_1 x)

Simplify the arbitrary constants:

Answer: |

| y(x) = -3/(x^4 + c_1 x)