# Thread: Having problem with this separation

1. ## Having problem with this separation

I'm pretty sure I'm doing something wrong.
the question is:
Formulate the general solution to the differential equation. (express in y^2 in terms of x)

$\displaystyle dy/dx = e^x/y + 2y$

This is what I did doing so far.

$\displaystyle dy/dx = e^x/y + 2y$

$\displaystyle dy/dx - 2y = e^x/y$
$\displaystyle dy/ydx = e^x +2$
$\displaystyle dy = y(e^x +2)dx$
$\displaystyle (1/y) dy = (e^x +2)dx$
$\displaystyle ln(y) = e^x + 2x$
$\displaystyle y = e^(e^x+2x)$
$\displaystyle y^2 =( e^(e^x+2x))^2$

the final answer just looks way too weird.

2. Originally Posted by konvos
I'm pretty sure I'm doing something wrong.
the question is:
Formulate the general solution to the differential equation. (express in y^2 in terms of x)

$\displaystyle dy/dx = e^x/y + 2y$

This is what I did doing so far.

$\displaystyle dy/dx = e^x/y + 2y$

$\displaystyle dy/dx - 2y = e^x/y$
$\displaystyle dy/ydx = e^x +2$
$\displaystyle dy = y(e^x +2)dx$
$\displaystyle (1/y) dy = (e^x +2)dx$
$\displaystyle ln(y) = e^x + 2x$
$\displaystyle y = e^(e^x+2x)$
$\displaystyle y^2 =( e^(e^x+2x))^2$

the final answer just looks way too weird.
First lets clean this up a bit

If we multiply by $\displaystyle y$ we get

$\displaystyle \displaystyle y\frac{dy}{dx}=e^{x}+2y^2$

Now make the substitution

$\displaystyle \displaystyle u=y^2 \implies \frac{du}{dx}=2y\frac{dy}{dx}$

Subbing this in gives

$\displaystyle \displaystyle \frac{1}{2}\frac{du}{dx}=e^{x}+2u \iff \frac{du}{dx}-4u=2e^{x}$

This equation can be solved via an integrating factor. Can you finish from here ?

3. Just as a further explanation of TheEmptySet's post: the original DE is Bernoulli, and hence a substitution of the form $\displaystyle u=y^{2}$ will render the equation linear in $\displaystyle u.$

4. Originally Posted by TheEmptySet
First lets clean this up a bit

If we multiply by $\displaystyle y$ we get

$\displaystyle \displaystyle y\frac{dy}{dx}=e^{x}+2y^2$

Now make the substitution

$\displaystyle \displaystyle u=y^2 \implies \frac{du}{dx}=2y\frac{dy}{dx}$

Subbing this in gives

$\displaystyle \displaystyle \frac{1}{2}\frac{du}{dx}=e^{x}+2u \iff \frac{du}{dx}-4u=2e^{x}$

This equation can be solved via an integrating factor. Can you finish from here ?
Thx for the help.
The Y^2 is getting my confused. The formula I'm applying is

$\displaystyle P(x) = -4, hence the derivative is -4x$
applying to both sides, I should get
$\displaystyle -4xy^2 = 2e^dx$
I just guessed that the y^2 would be squared, I tried looking for a example in my book and I couldn't find.
so the final answer is :
$\displaystyle y^2 = -2e^x/4x$ but like I said... I just guessed on the y^2, so i'm not sure

5. Originally Posted by konvos
Thx for the help.
The Y^2 is getting my confused. The formula I'm applying is
Why are you applying that formula, when it doesn't apply? That formula is the form of a first-order linear DE, but that's not what you start out with. If you will employ TheEmptySet's substitution, you will get a first-order linear DE in u, at which point you can employ the integrating factor method.

6. Originally Posted by Ackbeet
Why are you applying that formula, when it doesn't apply? That formula is the form of a first-order linear DE, but that's not what you start out with. If you will employ TheEmptySet's substitution, you will get a first-order linear DE in u, at which point you can employ the integrating factor method.
I'm confused by the substitution but if I substitute I'll get
$\displaystyle 2y(dy/dx) -4y^2= 2e^x$
$\displaystyle 2y(dy/dx - 2y) = 2e^x$
I'm not sure if the formula could be applied there.

7. Originally Posted by konvos
I'm confused by the substitution but if I substitute I'll get
$\displaystyle 2y(dy/dx) -4y^2= 2e^x$
$\displaystyle 2y(dy/dx - 2y) = 2e^x$
I'm not sure if the formula could be applied there.
You have not done the substition yet! If you look back in post #2 I did the substition. The NEW ODE has u as the dependant variable and x as the independant variable.

Use the formula on this equation.