#### 2.10   ODE No. 10

$y(x) f'(x)-f(x) f'(x)+y'(x)=0$ Mathematica : cpu = 0.0128871 (sec), leaf count = 18

$\left \{\left \{y(x)\to c_1 e^{-f(x)}+f(x)-1\right \}\right \}$ Maple : cpu = 0.01 (sec), leaf count = 15

$\left \{ y \left ( x \right ) =f \left ( x \right ) -1+{{\rm e}^{-f \left ( x \right ) }}{\it \_C1} \right \}$

Hand solution

\begin {equation} \frac {dy}{dx}+y\left ( x\right ) \frac {df}{dx}=f\left ( x\right ) \frac {df}{dx} \tag {1} \end {equation}

Integrating factor $$\mu =e^{\int \frac {df}{dx}dx}=e^{f}$$.   Therefore (1) becomes$\frac {d}{dx}\left ( e^{f}y\left ( x\right ) \right ) =e^{f}f\left ( x\right ) \frac {df}{dx}$ Integrating\begin {align*} e^{f}y\left ( x\right ) & =\int e^{f}f\left ( x\right ) \frac {df}{dx}dx+C\\ y\left ( x\right ) & =e^{-f}\int e^{f}fdf+e^{-f}C \end {align*}

But $$\int e^{f}fdf$$ is the same as $$\int e^{x}xdx$$ which by integration by parts gives $$e^{x}\left ( x-1\right )$$ or in terms of $$f$$, gives $$e^{f}\left ( f-1\right )$$. Hence the above becomes\begin {align*} y\left ( x\right ) & =e^{-f}\left ( e^{f}\left ( f-1\right ) \right ) +e^{-f}C\\ & =f-1+e^{-f}C \end {align*}