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#### paulmdrdo

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- May 13, 2013

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$\displaystyle (x^2+y^3+1)dx+x^4y^2dy=0$

I can't see any particular exact D.E form here. please help.

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- May 13, 2013

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$\displaystyle (x^2+y^3+1)dx+x^4y^2dy=0$

I can't see any particular exact D.E form here. please help.

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- May 13, 2013

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sorry MarkFL, I've already tried everything but still couldn't find something that's familiar to me.There is a special integrating factor that is a function of just $x$. Can you find it?

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If $M(x,y)\,dx+N(x,y)\,dy=0$ is neither separable nor linear, compute \(\displaystyle \frac{\partial M}{\partial y}\) and \(\displaystyle \frac{\partial N}{\partial x}\). If \(\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\), then the equation is exact. If it is not exact, consider:

(1) \(\displaystyle \frac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}\)

If (1) is a function of just $x$, then an integrating factor is given by:

\(\displaystyle \mu(x)=\exp\left(\int\frac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}\,dx \right)\)

If not, consider:

(2) \(\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}\)

If (2) is a function of just $y$, then an integrating factor is given by:

\(\displaystyle \mu(y)=\exp\left(\int\frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}\,dy \right)\)

Based on this, what do you conclude regarding the given problem?

- Jan 29, 2012

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MarkFL told you: "There is a special integrating factor that is a function of just x . Can you find it?"

Call that function v(t). Multiplying both sides of your differential equation by that gives

[tex]v(x)(x^2+ y^3+ 1)dx+ v(x)x^4y^2dy= 0[/tex]. The condition that this be "exact" would be [tex](v(x)(x^2+ y^3+ 1))_y= (v(x)x^4y^2)_x[/tex]

[tex]3v(x)y^2= v'(x)x^4y^2+ 4v(x)x^3y^3[/tex]

Now you can divide through by y to get the differential equation

[tex]3v= x^4v'+ 4x^3v[/tex]

[tex]x^4v'= (4x^3- 3)v[/tex]

A separable equation for v.

Call that function v(t). Multiplying both sides of your differential equation by that gives

[tex]v(x)(x^2+ y^3+ 1)dx+ v(x)x^4y^2dy= 0[/tex]. The condition that this be "exact" would be [tex](v(x)(x^2+ y^3+ 1))_y= (v(x)x^4y^2)_x[/tex]

[tex]3v(x)y^2= v'(x)x^4y^2+ 4v(x)x^3y^3[/tex]

Now you can divide through by y to get the differential equation

[tex]3v= x^4v'+ 4x^3v[/tex]

[tex]x^4v'= (4x^3- 3)v[/tex]

A separable equation for v.

Last edited:

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- May 13, 2013

- 386

I get an integrating factor which is just a function of x to be

$\displaystyle e^{-(x^{-3}+4\ln(x))}$ or $\displaystyle \frac{e^{-x^{-3}}}{x^4}$ now what am I going to do next?

Last edited:

Read HallsOfIvy's post.

I get an integrating factor which is just a function of x to be

$\displaystyle e^{-(x^{-3}+4\ln(x))}$ now what am I going to do next?

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- May 13, 2013

- 386

I get an integrating factor which is just a function of x to be

$\displaystyle e^{-(x^{-3}+4\ln(x))}$ or $\displaystyle \frac{e^{-x^{-3}}}{x^4}$ now what am I going to do next?

multiplying this integrating factor to my orig D.E i get,

$\displaystyle \left(\frac{e^{-x^{-3}}}{x^2}+\frac{e^{-x^{-3}}y^3}{x^4}+\frac{e^{-x^{-3}}}{x^4}\right)dx+e^{-x^{-3}}y^2dy$

now letting

$\displaystyle M=\left(\frac{e^{-x^{-3}}}{x^2}+\frac{e^{-x^{-3}}y^3}{x^4}+\frac{e^{-x^{-3}}}{x^4}\right)$

then,

$\displaystyle \frac{\partial M}{\partial y}=3x^{-4}y^2e^{-x^{-3}}$

and letting

$\displaystyle N=e^{-x^{-3}}y^2dy$

then,

$\displaystyle \frac{\partial N}{\partial x}=3x^{-4}y^2e^{-x^{-3}}$

now I have

$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

which tells us that I have an exact D.E

therefore there exist a Function $F(x,y)=c$ such that,

$\displaystyle\frac{\partial F}{\partial x}=M$ and $\displaystyle\frac{\partial F}{\partial y}=N$

now

$\displaystyle\frac{\partial F}{\partial x}=\left(\frac{e^{-x^{-3}}}{x^2}+\frac{e^{-x^{-3}}y^3}{x^4}+\frac{e^{-x^{-3}}}{x^4}\right)$

so, $\displaystyle F=\int \left(\frac{e^{-x^{-3}}}{x^2}+\frac{e^{-x^{-3}}y^3}{x^4}+\frac{e^{-x^{-3}}}{x^4}\right)\partial x$

can you help me continue with the integration? thanks!

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- May 13, 2013

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use integration by parts? Am I right?I would next consider writing:

\(\displaystyle F=\int e^{-x^{-3}}x^{-2}\,dx+\left(y^3+1 \right)\int e^{-x^{-3}}x^{-4}\,dx+g(y)\)

Now what can we do to each integral to get an appropriate differential for the obvious substitutions?

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Yes, eventually and on the first integral on the right only. But first we need to "fix" the two integrals so that the differentials we need in using an appropriate substitution will be present.use integration by parts? Am I right?

- Aug 7, 2013

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- May 13, 2013

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I don't know how to do that fixing of integrals.Yes, eventually and on the first integral on the right only. But first we need to "fix" the two integrals so that the differentials we need in using an appropriate substitution will be present.

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For an integral of the form:

\(\displaystyle \int e^{-x^{-3}}x^n\,dx\)

What would you begin by saying would be a good candidate for a substitution?

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- May 13, 2013

- 386

Sorry, still don't get it.

\(\displaystyle \int e^{-x^{-3}}x^n\,dx\)

What would you begin by saying would be a good candidate for a substitution?

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Don't you think the following substitution would be a good place to start?Sorry, still don't get it.

\(\displaystyle u=-x^{-3}\)

So, what does your differential then need to be?