Giải phương trình Bermolli : y' + =x

y(x) = -1/(x^2 + c_1 x)

Step-by-step 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:

y(x) = -1/(x^2 + c_1 x)

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