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let $C$ is give the constant ,if the function $f(x,y)$ such $$\dfrac{xf'_{x}}{f'_{y}}+\dfrac{yf'_{y}}{f'_{x}}+x+y=C$$

Find the all $f(x,y)$

I found this problem one solution: $$f(x,y)=\sqrt{x}+\sqrt{y}-C$$ is such it,because $$f'_{x}=\dfrac{1}{2\sqrt{x}},f'_{y}=\dfrac{1}{2\sqrt{y}}$$ so $$(xf'_{x}+yf'_{y})\left(\dfrac{1}{f'_{x}}+\dfrac{1}{f'_{y}}\right)=C$$

My question: this PDE Have other solution?if this solution is uniqueness,then How prove it?

this problem background is from this

show that: the length of the portion of any tangent line to the astroid $$f(x,y)=0$$ cut off by the coordinate axes is constant $C$,find $f(x,y)$

my idea: the tangent line is $$f'_{x}(X-x)+f'_{y}(Y-y)=0$$ so let $$x=0,\Longrightarrow Y=\dfrac{xf'_{x}}{f'_{y}}+y$$ let $$y=0\Longrightarrow X=\dfrac{yf'_{y}}{f'_{x}}+x$$ so $$X+Y=C\Longrightarrow \dfrac{xf'_{x}}{f'_{y}}+\dfrac{yf'_{y}}{f'_{x}}+x+y=C$$

then I can't solve this equation. Thank you for you help.

Mathlover
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math110
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1 Answers1

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Yes,if we write $$y=f(x)$$ then tangent line $$Y-f(x)=f'(x)[X-x]$$then $$X=-\dfrac{f(x)}{f'(x)}+x,Y=f(x)-xf'(x)$$ so $$f(x)-xf'(x)+x-\dfrac{f(x)}{f'(x)}=C$$ Both sides of the x derivation $$f'(x)-f'(x)-x[f'(x)]^2+1-\dfrac{[f'(x)]^2-f(x)f''(x)}{(f'(x))^2}=0$$ $$\Longrightarrow xf''(x)=\dfrac{f(x)f''(x)}{[f'(x)]^2}$$ case one $f''(x)=0$ then we have $$f(x)=kx+b$$ case 2: if $f''(x)\neq 0$,then we have $$[f'(x)]=\dfrac{f(x)}{x}\Longrightarrow \sqrt{f(x)}+\sqrt{x}=C$$ so $$\sqrt{y}+\sqrt{x}=C$$

math110
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