# Solve the separable differential equation: 2*sqrt(xy)*(dy/dx)=1

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You can rearrange your equation to separate the variables:

$$\begin{split} 2\sqrt{xy} {dy \over dx} &= 1 \\ 2\sqrt{y}dy&= \frac{1}{\sqrt{x}}dx \end{split}$$

Now integrate both side:

$$\begin{split} \int{2\sqrt{y}dy}&= \int{\frac{1}{\sqrt{x}}dx} \\ 2\frac{2}{3}y^{\frac{3}{2}} &= 2\sqrt{x} + C \\ y &= \bigg(\frac{3}{2}(\sqrt{x} + C_1)\bigg)^{\frac{2}{3}} \end{split}$$

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### user364321

Updated on November 24, 2022

• user364321 12 months

Solve the separable differential equation:

$2\sqrt{xy} {dy \over dx} = 1$

$x,y>0$

I understand the process. But I keep getting $C={4y^({2\over 3})\over 3}-2\sqrt{x}$ as my answer, when the book says the answer is $C={2y^({2\over 3})\over 3}-\sqrt{x}$. I just don't understand how/where the 2 is being factored out.

Rewrite the equation as $2\sqrt{y}dy=\dfrac{dx}{\sqrt{x}}$.
@user364321: Hint: You have $\displaystyle 2 \int \sqrt{y}~ dy = \int \dfrac{1}{\sqrt{x}}~dx$.
I integrated both sides but I keep getting $C={4y^({2\over 3})\over 3}-2\sqrt{x}$ as my answer. But the book says the answer is $C={2y^({2\over 3})\over 3}-\sqrt{x}$. I just don't understand how/where the 2 is being factored out.
I keep getting $C={4y^({2\over 3})\over 3}-2\sqrt{x}$ as my answer. But the book says the answer is $C={2y^({2\over 3})\over 3}-\sqrt{x}$. I just don't understand how/where the 2 is being factored out.
$C$ is just a constant so it doesn't matter if you factor out the 2. Look at the last two lines in my solution; note how I just replace $\frac{C}{2}$ with a new constant $C_1$.