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Solve the following differential equation:

$$y=x \cdot \frac{dy}{dx}+\frac{b}{\frac{dy}{dx}}$$ where $b$ is a real constant.

I tried solving it by taking $\frac{dy}{dx}=t$, then solving quadratic equation for roots. But this method gives complicated differential equation once I write expression for roots. Could someone suggest a better approach.

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    One obvious solution is $y=x+b$, or more generally, $y=cx+\frac bc$ – 2017-02-12
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    @HagenvonEitzen How can you directly say it? – 2017-02-12
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    Just plug it in – 2017-02-12

2 Answers 2

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This is a Clairaut equation.

Therefore, you can differentiate both sides w.r.t $x$ to obtain: $$\frac{dy}{dx}=x\frac{d^2 y}{dx^2}+\frac{dy}{dx}-\frac{b\cdot \frac{d^2 y}{dx^2}}{\left(\frac{dy}{dx}\right)^2}$$ Factoring gives: $$\frac{d^2 y}{dx^2}\left(x-\frac{b}{(\frac{dy}{dx})^2}\right)=0$$ Now, solve for the following separately: $$\frac{d^2 y}{dx^2}=0$$ And: $$x-\frac{b}{(\frac{dy}{dx})^2}=0$$

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i don't know if this helps.

let $m = \frac{dy}{dx}.$ now we can write your differential equation $y=x \cdot \frac{dy}{dx}+\frac{b}{\frac{dy}{dx}}$ as $$y = xm + \frac b m $$ solving this quadratic equation, you have $$2x\ m = y \pm \sqrt{y^2 - 4bx}\to 2x\frac{dy}{dx} = y\pm\sqrt{y^2 - 4bx} $$