Find the particular solution of the differential equation x^2 d y= (2 x y+y^2 ) d x given that y=1 when x=1

Step-by-step solution

The given differential equation is x² dy = (2xy + y²) dx , which can be written as dy dx = 2xy + y² x² . This is a homogeneous differential equation. Let y = vx , then dy dx = v + x dv dx . Substituting these into the equation: v + x dv dx = 2x(vx) + (vx)² x² = 2vx² + v²x² x² = 2v + v² . Rearranging the terms: x dv dx = 2v + v² - v = v² + v . Separating the variables: dv v² + v = dx x , which is dv v(v+1) = dx x . Using partial fractions: ( 1 v - 1 v+1 ) dv = dx x . Integrating both sides: |v| - |v+1| = |x| + C , w