# Having problem with this separation

• Feb 5th 2011, 09:30 AM
konvos
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)

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

This is what I did doing so far.

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

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

$y^2 =( e^(e^x+2x))^2
$

the final answer just looks way too weird.
• Feb 5th 2011, 09:44 AM
TheEmptySet
Quote:

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)

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

This is what I did doing so far.

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

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

$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 $y$ we get

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

Now make the substitution

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

Subbing this in gives

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

This equation can be solved via an integrating factor. Can you finish from here ?
• Feb 5th 2011, 10:23 AM
Ackbeet
Just as a further explanation of TheEmptySet's post: the original DE is Bernoulli, and hence a substitution of the form $u=y^{2}$ will render the equation linear in $u.$
• Feb 5th 2011, 10:26 AM
konvos
Quote:

Originally Posted by TheEmptySet
First lets clean this up a bit

If we multiply by $y$ we get

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

Now make the substitution

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

Subbing this in gives

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

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

$P(x) = -4, hence the derivative is -4x$
applying to both sides, I should get
$-4xy^2 = 2e^dx$
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 :
$y^2 = -2e^x/4x$ but like I said... I just guessed on the y^2, so i'm not sure
• Feb 5th 2011, 10:30 AM
Ackbeet
Quote:

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.
• Feb 5th 2011, 10:42 AM
konvos
Quote:

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
$2y(dy/dx) -4y^2= 2e^x$
$2y(dy/dx - 2y) = 2e^x$
I'm not sure if the formula could be applied there.
• Feb 5th 2011, 10:54 AM
TheEmptySet
Quote:

Originally Posted by konvos
I'm confused by the substitution but if I substitute I'll get
$2y(dy/dx) -4y^2= 2e^x$
$2y(dy/dx - 2y) = 2e^x$
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.