$$x\sqrt{1+y} + y\sqrt{1+x} = 0$$
Please tell me where I went wrong. Why I am not getting correct answer ?
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I suggest you write the original equation as: $$x\sqrt{1+y}=-y \sqrt{1+x}$$ then square both sides and find explicit expression for $y(x)$ – Yuriy S Jun 10 '16 at 06:08
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1You might not even have gone wrong, you can still simplify the equation by replacing $\sqrt{1+y}/\sqrt{1+x}$ by $-y/x$. – Asvin Jun 10 '16 at 06:09
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I do not see anything wrong... maybe u should use the given equation again in ur result! – Jasser Jun 10 '16 at 06:11
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There appears nothing wrong with your work, It is only that you have chosen a way that does gives you desired form of answer. – Shikhar S. Maheshwari Jun 10 '16 at 06:38
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Some other posts about the same problem: Differentiate $x \sqrt{1+y}+y \sqrt{1+x}=0$ and If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$. – Martin Sleziak Apr 16 '18 at 00:59
2 Answers
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There is nothing wrong. Put the value of $y$ to get your result.
However, a simpler approach:
$$x\sqrt{1+y} + y\sqrt{1+x} = 0$$ $$x\sqrt{1+y} = - y\sqrt{1+x}$$ Squarring both sides, we get $$x^2(1+y) = y^2(1+x)$$ $$x^2(1+y) - y^2(1+x)=0$$ $$(x-y)(x+y+xy)=0$$
So either $x-y=0$ or $x+y+xy=0$.
Now if, $x-y=0$, then we have $x=y$. But this does not satisfy the equation $x\sqrt{1+y} + y\sqrt{1+x} = 0$ for all $x,y \in \mathbb{R}$.
Hence required solution is $x+y+xy=0$.
Thus we can write that
$$1+\frac{dy}{dx}+y+x\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{-y-1}{1+x}=\frac{\frac{x}{1+x}-1}{1+x}=-\frac{1}{(1+x)^2}$$
SchrodingersCat
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I see nothing wrong with your work....
Use the given equation in question in your result to simplify it further to get the required answer.
Jasser
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