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Obtain the complete primitive and singular solutions of the differential equation: $$y-x\frac{dy}{dx}-(\frac{dy}{dx})^2=0$$

Could someone please explain to me how the equation can be solved?

Does it need to be translated to $$y''+xy'-y=0$$

and which type of method do you need to use to solve it?

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    Do you feel the difference between : $y''$ and $(y')^2$ ? – 2017-01-13
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    For example : $(x^2)'' = 2$ and $((x^2)' )^2 = 4x^2$ – 2017-01-13
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    solve the equation for $$\frac{dy}{dx}$$ – 2017-01-13
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    If its the first differential equation which is the original problem then I think it would be easier if you differentiate the equation which gives you $$-y''(2y'+x)=0$$ from here you get two differential equations to solve, which are pretty easy to do. – 2017-01-13
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    Thank you all for your answers. You have been very helpful. – 2017-01-13

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Let us rewrite the equation as $$y-xy'-(y')^2=0$$ Differentiating both sides with respect to $x$, we get: $$ y'-y'-xy''-2y'y''=0$$ or equivalently: $$ xy''+2y'y''=y''(x+2y')=0$$ Thus $y''=0$ or $x+2y'=0$. If $y''=0$, then $y=c_1x+c_2$. If $x+2y'=0$, then $y'=\frac{-x}{2}$ and $y=-\frac{x^2}{4}+c_3$.

Therefore, the solution to your differential equation is $y=-\frac{x^2}{4}+c_3$ or $y=c_1x+c_2$.

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    You know even more by inserting the first derivative into the original equation. In the first case $y''=0$, one gets $y=cx+c^2$ with $c=y'$ and in the second $y=x·(-\frac x2)+(-\frac x2)^2=-\frac{x^2}4$. See Clairaut differential equation. – 2017-01-13