solution of the equationx dy/dx + y=y2logx is given by

The given differential equation is a Bernoulli's equation. The general form of a Bernoulli's equation is dy/dx + P(x)y = Q(x)y^n.

In the given equation, x dy/dx + y = y^2 logx, we can rewrite it as dy/dx + y/x = y^2 logx.

Here, P(x) = 1/x, Q(x) = logx, and n = 2.

The solution to Bernoulli's equation is obtained by substituting y = v^(1-n), where v is a new dependent variable.

So, let's substitute y = v^(1-2) = 1/v.

Then, dy/dx = -1/v^2 dv/dx.

Substituting these into the original equation, we get -1/v^2 dv/dx + 1/(xv) = logx/v.

This simplifies to -dv/dx + 1/x dv = -logx.

This is a linear differential equation in v and x, which can be solved using an integrating factor.

The integrating factor is e^(∫P(x) dx) = e^(∫(1/x) dx) = e^log|x| = x.

Multiplying the equation by the integrating factor, we get -x dv/dx + dv = -x logx.

This can be written as d/dx(xv) = -x logx.

Integrating both sides, we get xv = -∫x logx dx.

The integral on the right side can be solved using integration by parts, with u = logx and dv = x dx.

This gives ∫x logx dx = 1/2 x^2 logx - ∫1/2 x dx = 1/2 x^2 logx - 1/4 x^2.

Substituting this back into the equation, we get xv = -1/2 x^2 logx + 1/4 x^2.

Finally, substituting back v = 1/y, we get xy = -1/2 x^2 logx + 1/4 x^2.

Solving for y, we get y = -1/2 x logx + 1/4 x.

So, the solution to the given differential equation is y = -1/2 x logx + 1/4 x.