making equation exact

• Feb 9th 2011, 01:51 PM
duaneg37
making equation exact
Solve $(x^{2}+y^{2}-5)dx=(y+xy)dy$
$y(0)=1$

I have tried finding a factor to make this equation exact, and I get $(x+1)^{-3}$. I then integrate by parts with respect to x and get $\,{\frac {-{x}^{2}-{y}^{2}+5}{2 \left( x+1 \right) ^{2}}}+\ln
\left( x+1 \right) + \left( x+1 \right) ^{-1}+\rho \left( y \right) =
{\frac {{x}^{2}+{y}^{2}-5}{ \left( x+1 \right) ^{3}}}

$

I feel like I'm not getting anywhere because nothing cancels like it should. Can someone please help me? Thanks!
• Feb 9th 2011, 03:23 PM
General
$(x^{2}+y^{2}-5)dx-y(1+x)dy=0$

Multiplying by the integrating factor

$
\displaystyle \dfrac{x^2+y^2-5}{(x+1)^3} dx + \left( \dfrac{-y}{(x+1)^2} \right) dy = 0
$

The last equation is exact. Its solution of the form $F(x,y)=C$ where :

$F_x=\dfrac{x^2+y^2-5}{(x+1)^3}$ ... (1)

$F_y=\dfrac{-y}{(x+1)^2}$ ... (2)

I think you integrate (1) with respect to x. This is too complicated .

Try to integrate (2) with respect to y and put p(x) as the constant of integration.
• Feb 9th 2011, 06:47 PM
duaneg37
$\frac{\partial{f}}{\partial{y}}={\frac {-{y}{\it }}{ \left( x+1 \right) ^{2}}}dy$ Now I integrate w.r.t.y

$f( x,y) ={\frac {-{y}^{2}}{ \left2( x+1 \right) ^{2}}
}+p \left( x \right)$
Now I take partial w.r.t.x

$\frac{\partial{f}}{\partial{x}}={\frac {-{y}^{2}}{4(x+1)}}+\mbox p' \left( x
\right) ={\frac { \left( {x}^{2}+{y}^{2}-5 \right) {\it }}{ \left(
x+1 \right) ^{3}}}$
I thought things would cancel if it was exact. Can anyone tell me what I'm doing wrong? Thanks!
• Feb 9th 2011, 08:29 PM
duaneg37
${{\rm e}^{\int \!{\frac {M_{{y}}-N_{{x}}}{N}}{dx}}}$

This is how I get $(x+1)^{-3}$

How do you get x+1?
• Feb 9th 2011, 09:01 PM
topsquark
Quote:

Originally Posted by duaneg37
${{\rm e}^{\int \!{\frac {M_{{y}}-N_{{x}}}{N}}{dx}}}$

This is how I get $(x+1)^{-3}$

How do you get x+1?

Okay, I see my mistake. Hang on.

-Dan
• Feb 9th 2011, 09:31 PM
topsquark
Quote:

Originally Posted by General
$(x^{2}+y^{2}-5)dx-y(1+x)dy=0$

Multiplying by the integrating factor

$
\displaystyle \dfrac{x^2+y^2-5}{(x+1)^3} dx + \left( \dfrac{-y}{(x+1)^2} \right) dy = 0
$

The last equation is exact. Its solution of the form $F(x,y)=C$ where :

$F_x=\dfrac{x^2+y^2-5}{(x+1)^3}$ ... (1)

$F_y=\dfrac{-y}{(x+1)^2}$ ... (2)

I think you integrate (1) with respect to x. This is too complicated .

Try to integrate (2) with respect to y and put p(x) as the constant of integration.

Okay, now that I've fixed my little problem with the minus signs...
(1) isn't difficult to integrate, but it is tedious. You should get
$F(x, y) = ln|x + 1| + \frac{4x -y^2 + 8}{2(x + 1)^2} + \phi (y)$

What I've always done from here is to integrate (2). This is an easy one. Then just match up common terms. It saves from having to work with the derivative of what you called p(x).

-Dan
• Feb 9th 2011, 10:08 PM
duaneg37
$f(x,y)=\ln \left( x+1 \right) +\,{\frac {4\,x-{y}^{2}+8}{ 2\left( x+1
\right) ^{2}}}-\frac{7}{2}=1$

Does this look right? What technique did you use to integrate $F_{{x}}$?
• Feb 10th 2011, 12:06 AM
topsquark
$F(x, y) = \int \frac{x^2 + y^2 - 5}{(x + 1)^3} dx + \phi(y)$

So we need to evaluate the integral
$\int \frac{x^2 + y^2 - 5}{(x + 1)^3} dx$

$= \int \frac{x^2}{(x + 1)^3} dx + \int \frac{y^2 - 5}{(x + 1)^3}~dx$

The second integral we can evaluate directly:
$= \int \frac{x^2}{(x + 1)^3} dx - \frac{y^2 - 5}{2(x + 1)^2}$

For the first integral use a = x + 1:

$= \int \frac{(a - 1)^2}{a^3}da - \frac{y^2 - 5}{2(x + 1)^2}$

$= \int \left ( \frac{1}{a} - \frac{2}{a^2} + \frac{1}{a^3} \right ) da - \frac{y^2 - 5}{2(x + 1)^2}$

You can fill in the rest from there.

-Dan
• Feb 10th 2011, 12:14 AM
topsquark
Quote:

Originally Posted by duaneg37
$f(x,y)=\ln \left( x+1 \right) +\,{\frac {4\,x-{y}^{2}+8}{ 2\left( x+1
\right) ^{2}}}-\frac{7}{2}=1$

Does this look right? What technique did you use to integrate $F_{{x}}$?

What did you get when you integrated that second function?
$F(x, y) = -\int \dfrac{y}{(x + 1)^2}dy + \omega (x) = -\dfrac{1}{(x + 1)^2} \int y~dy + \omega (x)$

$= -\dfrac{y^2}{2(x + 1)^2} + \omega (x)$

Now just compare the two forms.

By the way, please note the distinction between $ln(x + 1)$ and the correct $ln|x + 1|$

-Dan
• Feb 10th 2011, 06:40 AM
General
Quote:

Originally Posted by duaneg37
$\frac{\partial{f}}{\partial{y}}={\frac {-{y}{\it }}{ \left( x+1 \right) ^{2}}}dy$ Now I integrate w.r.t.y

$f( x,y) ={\frac {-{y}^{2}}{ \left2( x+1 \right) ^{2}}
}+p \left( x \right)$
Now I take partial w.r.t.x

$\frac{\partial{f}}{\partial{x}}={\frac {-{y}^{2}}{4(x+1)}}+\mbox p' \left( x
\right) ={\frac { \left( {x}^{2}+{y}^{2}-5 \right) {\it }}{ \left(
x+1 \right) ^{3}}}$
I thought things would cancel if it was exact. Can anyone tell me what I'm doing wrong? Thanks!

Your $\displaystyle \frac{\partial{f}}{\partial{x}}$ is wrong.

You should get : $\displaystyle \dfrac{y^2}{(x+1)^3}+p'(x)$ ... (1)

And , from the original equation since M= $F_x$ we have $\displaystyle \frac{\partial{f}}{\partial{x}}=\dfrac{x^2+y^2-5}{(x+1)^3}$

Or : $\displaystyle \frac{\partial{f}}{\partial{x}}=\dfrac{y^2}{(x+1)^ 3}+\dfrac{x^2-5}{(x+1)^3}$ ... (2)

Comparing (1) & (2) gives : $\displaystyle p'(x)=\dfrac{x^2-5}{(x+1)^3}$

I think you can continue ...