MHB Differential Equation of Unknown Type

[paulmdrdo1]

can you help me solve this,

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

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

 

[MarkFL]

There is a special integrating factor that is a function of just $x$. Can you find it?

 

[paulmdrdo1]

There is a special integrating factor that is a function of just $x$. Can you find it?

sorry MarkFL, I've already tried everything but still couldn't find something that's familiar to me. :(

 

[MarkFL]

This is what I was taught as a student:

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

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

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

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

If not, consider:

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

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

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

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

 

[HallsofIvy]

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

Now you can divide through by y to get the differential equation
3v= x^4v'+ 4x^3v

x^4v'= (4x^3- 3)v

A separable equation for v.

 

[paulmdrdo1]

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?

 

[Prove It]

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?

Read HallsOfIvy's post.

 

[paulmdrdo1]

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!

 

[MarkFL]

I would next consider writing:

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

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

 

[paulmdrdo1]

I would next consider writing:

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

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

use integration by parts? Am I right?

 

[MarkFL]

use integration by parts? Am I right?

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.

 

[paulmdrdo]

:) :)

 

[paulmdrdo1]

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.

I don't know how to do that fixing of integrals.

 

[MarkFL]

For an integral of the form:

$$\int e^{-x^{-3}}x^n\,dx$$

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

 

[paulmdrdo1]

For an integral of the form:

$$\int e^{-x^{-3}}x^n\,dx$$

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

Sorry, still don't get it.:(

 

[MarkFL]

Sorry, still don't get it.:(

Don't you think the following substitution would be a good place to start?

$$u=-x^{-3}$$

So, what does your differential then need to be?