# Thread: a differential equation

1. ## a differential equation

would anyone be able to show us how you would find the general solution of this differential equation in implicit form for

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y^{1/2}(2e^{2x}-5)}{3(e^{2x}-5x)^{2/3}}$ $\displaystyle (y>0)$

2. ## Re: a differential equation

Pretty close to trivial, isn't it? That's clearly a separable equation:
$\displaystyle y^{-1/2}dy= \frac{2}{3}\frac{\left(e^{2x}- 5x\right)^{-2/3}}{2e^{2x}- 5}dx$

Integrate. (Do you recognize that $\displaystyle (e^{2x}- 5x)'= 2e^{2x}- 5$?)

3. ## Re: a differential equation

yes I see that. I have only started calculus a few weeks ago so I do apologize. It is the y and what is ment by implicit form that is throwing me

4. ## Re: a differential equation

"Implicit form" simply means that you don't have to solve for y. Clearly when you integrate the left side you have something involving $\displaystyle y^{1/2}$. Leave it in that form.

5. ## Re: a differential equation

So would the correct answer be

$\displaystyle \frac{y^{1/2}}{1/2}=6(2e^{2x}-5x)^{1/3}+c$

6. ## Re: a differential equation

Originally Posted by Orlando
would anyone be able to show us how you would find the general solution of this differential equation in implicit form for

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y^{1/2}(2e^{2x}-5)}{3(e^{2x}-5x)^{2/3}}$ $\displaystyle (y>0)$
$\displaystyle \int \frac{dy}{\sqrt y} = \int \frac{2(2e^{2x}-5)}{3(e^{2x}-5x)^{2/3}}\,dx$

Hence :

$\displaystyle 2\sqrt y=2\sqrt[3]{e^{2x}-5x} + C \Rightarrow \sqrt y=\sqrt[3]{e^{2x}-5x} + C$