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Find the solution of the differential equation $(\frac{dy}{dx})^2-x(\frac{dy}{dx})+y=0$.

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1 Answer

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Assuming ODE $(y')^2-xy'+y=0$, I tried the following:

Differentiate the equation to get $$2y'y''-y'-xy''+y'=0 \implies (2y'-x)y''=0.$$

So $y''=0\implies y =ax+b$ or $2y'-x=0\implies y=0.25x^2+c$.

Plug these solutions into the original ODE: $a^2-xa+ax+b=0\implies b =-a^2 \implies y = ax-a^2$ or $(0.5x)^2-x(0.5x)+(0.25x^2+c)=0 \implies c=0 \implies y=0.25x^2$

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An alternative approach would be to solve for $y'$ using the quadratic formula $$y'_{1/2}=\frac{x\pm \sqrt{x^2-4y}}{2}.$$ But the resulting ODE is nonlinear.

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