# Thread: making equation exact

1. ## making equation exact

Solve $\displaystyle (x^{2}+y^{2}-5)dx=(y+xy)dy$
$\displaystyle y(0)=1$

I have tried finding a factor to make this equation exact, and I get $\displaystyle (x+1)^{-3}$. I then integrate by parts with respect to x and get $\displaystyle \,{\frac {-{x}^{2}-{y}^{2}+5}{2 \left( x+1 \right) ^{2}}}+\ln \left( x+1 \right) + \left( x+1 \right) ^{-1}+\rho \left( y \right) = {\frac {{x}^{2}+{y}^{2}-5}{ \left( x+1 \right) ^{3}}}$
I feel like I'm not getting anywhere because nothing cancels like it should. Can someone please help me? Thanks!

2. Rewrite your differential as
$\displaystyle (x^{2}+y^{2}-5)dx-y(1+x)dy=0$

Multiplying by the integrating factor

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

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

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

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

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

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

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

4. $\displaystyle {{\rm e}^{\int \!{\frac {M_{{y}}-N_{{x}}}{N}}{dx}}}$

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

How do you get x+1?

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

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

How do you get x+1?
Okay, I see my mistake. Hang on.

-Dan

6. Originally Posted by General
Rewrite your differential as
$\displaystyle (x^{2}+y^{2}-5)dx-y(1+x)dy=0$

Multiplying by the integrating factor

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

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

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

$\displaystyle F_y=\dfrac{-y}{(x+1)^2}$ ... (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
$\displaystyle F(x, y) = ln|x + 1| + \frac{4x -y^2 + 8}{2(x + 1)^2} + \phi (y)$

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

7. $\displaystyle f(x,y)=\ln \left( x+1 \right) +\,{\frac {4\,x-{y}^{2}+8}{ 2\left( x+1 \right) ^{2}}}-\frac{7}{2}=1$

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

8. $\displaystyle F(x, y) = \int \frac{x^2 + y^2 - 5}{(x + 1)^3} dx + \phi(y)$

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

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

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

For the first integral use a = x + 1:

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

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

You can fill in the rest from there.

-Dan

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

Does this look right? What technique did you use to integrate $\displaystyle F_{{x}}$?
What did you get when you integrated that second function?
$\displaystyle F(x, y) = -\int \dfrac{y}{(x + 1)^2}dy + \omega (x) = -\dfrac{1}{(x + 1)^2} \int y~dy + \omega (x)$

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

Now just compare the two forms.

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

-Dan

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

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

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

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

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

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

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

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

I think you can continue ...