# solve first grade ecuation with bernoulli dy/dx = (y^2+2xy)/x^2general ecuation and particular when y(1)=1

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### 1 Answer

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

`=y^2/x^2+(2xy)/x^2`

`=(y/x)^2+2(y/x)`

Now let v = y/x. Then y = vx and y' = v'x+v

`dy/dx=v^2+2v=(dv)/dx*x+v`

`v^2+v=x(dv)/dx`

`1/xdx=1/(v^2+v)dv`

`lnx = lnv-ln(v+1)+lnc`

`lnx = ln((vc)/(v+1))`

`x=(vc)/(v+1)`

`x = cy/x*1/(y/x+1)=cy/x*x/(y+x)`

`xy+x^2 = cy`

`cy-xy=x^2`

`y(c-x)=x^2`

y = `x^2/(c-x)`

Now if y(1) = 1, then 1 = 1/(c-1) so c = 2.

`y=x^2/(2-x)`