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

calculus ordinary-differential-equations

4,176

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:

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

4,176

user364321

Updated on November 24, 2022

Comments

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.
- Admin about 7 years
  
  Rewrite the equation as $2\sqrt{y}dy=\dfrac{dx}{\sqrt{x}}$.
- Moo about 7 years
  
  @user364321: Hint: You have $\displaystyle 2 \int \sqrt{y}~ dy = \int \dfrac{1}{\sqrt{x}}~dx$.
- user364321 about 7 years
  
  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.
user364321 about 7 years

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.
gowrath about 7 years

$C$ 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 $C_1$.
user364321 about 7 years

Oh, ok. I understand now. Thank you!