# Solve the differential quation: dy/dx = -(2xy^2+1)/(2x^2y)

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### 2 Answers

We have to solve dy/dx = -(2xy^2+1)/(2x^2y)

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

=> (2x^2y) dy = (-2xy^2 - 1) dx

Integrate both the sides

=> 2x^2y^2/2 = -2y^2x^2/2 - x +C

=> x^2y^2 = -x^2y^2 - x +C

=> 2x^2y^2 = -x +C

Moving all terms to one side, we get:

**The required equation is 2x^2y^2 + x + C = 0**

Given the differential equation:

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

First we will cross multiply.

==> (2x^2 y ) dy = -(2xy^2 + 1) dx

Now we will integrate both sides.

==> Int 2x^2y dy = Int -(2xy^2 + 1) dx

==> 2x^2 y^2 /2 = -(2x^2*y^2/2 + x) + C

==> x^2 y^2 = - x^2*y^2 - x + C

==> 2x^2 *y^2 + x = C

==> Then the equation is given by :

**2x^2y^2 + x + C = 0**