The second order differential equation is given as:

\(xy''+x(y')^2-y'=0\)

isolated y''

\(y''=-\frac{x(y')^2-y'}{x}\)

Substitute y'=v and v is the function of x:

\(y''=-\frac{x(y')^2-y'}{x}\)

\(y'=v\)

\(v'=-\frac{x(v)^2-v}{x}\Rightarrow\frac{dv}{dx}=-\frac{x(v)^2-v}{x}\)

Now, the equation is of thr form:

A first order Bernoulli IDE has the form of

\(y'+p(x)y=q(x)y^n\)

\(v'-\frac{1}{x}v=-v^2\)

The general solution is obtained by substituting \(u=v^(1-n)\)

solving \frac{1}{1-n}u'+p(x)u=q(x)\)

put \(u=v^{-1}\)

\(-u'-\frac{u}{x}=-1\Rightarrow u'+\frac{u}{x}=1\)

Rewrite the first irderliner ODE and solve the equation:

\(u'+\frac{u}{x}=1\)

Rewrite in the form of a first order linear ODE.

\(u'(x)+p(x)\cdot u=q(x)\)

\(p(x)=\frac{1}{x},\ q(x)=1\)

\(I.F.=e^{\int\frac{1}{x}dx}=x\)

\(u(x)\cdot x=\int1\cdot u(x)dx=\int xdx=\frac{x^2}{2}+c_1\)

\(u(x)=\frac{x}{2}+\frac{c_1}{x}=\frac{x^2+2c_1}{2x}\) Let \(2c_1=c\)

\(v(x)=\frac{1}{u(x)}=\frac{2x}{x^2+2c_1}\)

Substitute back the values of \(v(x)=y'\) and obtain the solution:

Substitute back \(v=y'\)

\(v(x)=y'(x)=\frac{2x}{x^2+c}\)

\(\frac{dy}{dx}=\frac{2x}{x^2+c}\Rightarrow dy=\frac{2x}{x^2+c}dx\)

integrating both sides:

\(\int dy=\int\frac{2x}{x^2+c}dx\)

\(y+c_2=\log(x^2+c)\)

\(y=\log(x^2+c)-c_2\)

\(xy''+x(y')^2-y'=0\)

isolated y''

\(y''=-\frac{x(y')^2-y'}{x}\)

Substitute y'=v and v is the function of x:

\(y''=-\frac{x(y')^2-y'}{x}\)

\(y'=v\)

\(v'=-\frac{x(v)^2-v}{x}\Rightarrow\frac{dv}{dx}=-\frac{x(v)^2-v}{x}\)

Now, the equation is of thr form:

A first order Bernoulli IDE has the form of

\(y'+p(x)y=q(x)y^n\)

\(v'-\frac{1}{x}v=-v^2\)

The general solution is obtained by substituting \(u=v^(1-n)\)

solving \frac{1}{1-n}u'+p(x)u=q(x)\)

put \(u=v^{-1}\)

\(-u'-\frac{u}{x}=-1\Rightarrow u'+\frac{u}{x}=1\)

Rewrite the first irderliner ODE and solve the equation:

\(u'+\frac{u}{x}=1\)

Rewrite in the form of a first order linear ODE.

\(u'(x)+p(x)\cdot u=q(x)\)

\(p(x)=\frac{1}{x},\ q(x)=1\)

\(I.F.=e^{\int\frac{1}{x}dx}=x\)

\(u(x)\cdot x=\int1\cdot u(x)dx=\int xdx=\frac{x^2}{2}+c_1\)

\(u(x)=\frac{x}{2}+\frac{c_1}{x}=\frac{x^2+2c_1}{2x}\) Let \(2c_1=c\)

\(v(x)=\frac{1}{u(x)}=\frac{2x}{x^2+2c_1}\)

Substitute back the values of \(v(x)=y'\) and obtain the solution:

Substitute back \(v=y'\)

\(v(x)=y'(x)=\frac{2x}{x^2+c}\)

\(\frac{dy}{dx}=\frac{2x}{x^2+c}\Rightarrow dy=\frac{2x}{x^2+c}dx\)

integrating both sides:

\(\int dy=\int\frac{2x}{x^2+c}dx\)

\(y+c_2=\log(x^2+c)\)

\(y=\log(x^2+c)-c_2\)