\[ f(x) \left (y(x)^2-x^2\right )+x y'(x)=0 \] ✗ Mathematica : cpu = 12.552 (sec), leaf count = 0
, could not solve
DSolve[f[x]*(-x^2 + y[x]^2) + x*Derivative[1][y][x] == 0, y[x], x]
✗ Maple : cpu = 0. (sec), leaf count = 0
, could not solve
dsolve(x*diff(y(x),x)+f(x)*(y(x)^2-x^2) = 0,y(x))
Hand solution
\(xy^{\prime }+axy^{2}+2y+bx=0\)This is Riccati non-linear first order. Converting it to standard form\begin {align} xy^{\prime }+f\relax (x) \left (y^{2}-x^{2}\right ) -y & =0\nonumber \\ xy^{\prime } & =-f\left (y^{2}-x^{2}\right ) +y\nonumber \\ y^{\prime } & =-\frac {f}{x}y^{2}+fx+\frac {1}{x}y\tag {1} \end {align}
This is Riccati non-linear first order oder. There are two particular solutions \(y_{p}=\pm x\). Using \(y_{p}=x\), then using the transformation \(y=y_{p}+\frac {1}{u}\), gives \(y^{\prime }=1-\frac {u^{\prime }}{u^{2}}\) and (1) becomes\begin {align*} 1-\frac {u^{\prime }}{u^{2}} & =-\frac {f}{x}\left (x+\frac {1}{u}\right ) ^{2}+fx+\frac {1}{x}\left (x+\frac {1}{u}\right ) \\ & =-\frac {f}{x}\left (x^{2}+\frac {1}{u^{2}}+2\frac {x}{u}\right ) +fx+\left ( 1+\frac {1}{ux}\right ) \\ & =-fx-\frac {f}{x}\frac {1}{u^{2}}-2\frac {f}{u}+fx+1+\frac {1}{ux}\\ & =-\frac {f}{x}\frac {1}{u^{2}}-2\frac {f}{u}+1+\frac {1}{ux} \end {align*}
Hence\begin {align*} u^{2}-u^{\prime } & =-\frac {f}{x}-2fu+u^{2}+\frac {u}{x}\\ -u^{\prime } & =-\frac {f}{x}-2fu+\frac {u}{x}\\ -u^{\prime }-\frac {u}{x}+2fu & =\frac {-f}{x}\\ u^{\prime }+\frac {u}{x}-2fu & =\frac {-f}{x}\\ u^{\prime }+u\left (\frac {1}{x}-2f\right ) & =\frac {-f}{x} \end {align*}
Integrating factor is \(\mu =e^{\int \frac {1}{x}-2f}=e^{\ln x}e^{-2\int fdx}=xe^{-2\int fdx}\), hence\begin {align*} d\left (\mu u\right ) & =-\mu \frac {f}{x}\\ d\left (xe^{-2\int fdx}u\right ) & =-\left (xe^{-2\int fdx}\right ) \frac {f}{x}\\ d\left (xe^{-2\int fdx}u\right ) & =-f\left (e^{-2\int fdx}\right ) \end {align*}
Integrating\begin {align*} xe^{-2\int fdx}u & =-\int f\left (e^{-2\int fdx}\right ) +C\\ u & =-\frac {1}{x}e^{2\int fdx}\int f\left (e^{-2\int fdx}\right ) +C\frac {1}{x}e^{2\int fdx} \end {align*}
Since \(u=\frac {1}{y}\) then\begin {align*} y & =\frac {1}{-\frac {1}{x}e^{2\int fdx}\int f\left (e^{-2\int fdx}\right ) +C\frac {1}{x}e^{2\int fdx}}\\ & =\frac {xe^{-2\int fdx}}{-\int fe^{-2\int fdx}dx+C} \end {align*}
Verification (Maple does not verify it, need to look more into this)