# Thread: Linear , homogenous, separation and change of variables

1. ## Re: Linear , homogenous, separation and change of variables

Can you show whether it is exact or not?

2. ## Re: Linear , homogenous, separation and change of variables

My = x + 1
Nx = 6x - 1

Is this right?

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3. ## Re: Linear , homogenous, separation and change of variables

I mean not exact right..?

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4. ## Re: Linear , homogenous, separation and change of variables

(x+y)dx + (3x^2 - 1)dy = 0

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5. ## Re: Linear , homogenous, separation and change of variables

The equation is not exact, but check your partial differentiation...

6. ## Re: Linear , homogenous, separation and change of variables

Nx= 6x?

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7. ## Re: Linear , homogenous, separation and change of variables

$\displaystyle M_y(x,y)=1$

$\displaystyle N_y(x,y)=6x$

Since these are not equal, the equation is not exact. So, what should we do to seek an integrating factor?

Just to be clear, the equation is actually:

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

Correct?

8. ## Re: Linear , homogenous, separation and change of variables

Yes. Its

(x+y)dx + (3x^2 - 1)dy = 0

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9. ## Re: Linear , homogenous, separation and change of variables

Originally Posted by romeobernard
Yes. Its

(x+y)dx + (3x^2 - 1)dy = 0
Okay, good...that makes finding an integrating factor possible. Can you find such a factor?

10. ## Re: Linear , homogenous, separation and change of variables

I think there is no way we can find integrating factor base on the standard linear form.

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11. ## Re: Linear , homogenous, separation and change of variables

Originally Posted by romeobernard
I think there is no way we can find integrating factor base on the standard linear form.
According to the theory I was taught as a student, we should try:

$\displaystyle \mu(x)=\exp\left(\int \frac{1-6x}{3x^2-1}\,dx\right)=\,?$

12. ## Re: Linear , homogenous, separation and change of variables

Originally Posted by MarkFL
According to the theory I was taught as a student, we should try:

$\displaystyle \mu(x)=\exp\left(\int \frac{1-6x}{3x^2-1}\,dx\right)=\,?$
Ah..yes i see it its the IF for non exact equation.

But the next step is using a partial fraction differentiation right?

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13. ## Re: Linear , homogenous, separation and change of variables

Yes, partial fraction decomposition on the integrand would be a good idea, so you can get the integral in terms of natural logs.

14. ## Re: Linear , homogenous, separation and change of variables

(x+2y-1)dx - (2x+y-5)dy = 0

Thanks.

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15. ## Re: Linear , homogenous, separation and change of variables

What I would do first is solve the system:

$\displaystyle x+2y-1=0$

$\displaystyle 2x+y-5=0$

From which we obtain:

$\displaystyle (x,y)=(3,-1)$

And so let's let:

$\displaystyle x=u+3\implies dx=du$

$\displaystyle y=v-1\implies dy=dv$

And the ODE becomes:

$\displaystyle ((u+3)+2(v-1)-1)dx-(2(u+3)+(v-1)-5)dy=0$

$\displaystyle (u+2v)du-(2u+v)dv=0$

$\displaystyle \frac{dv}{du}=\frac{u+2v}{2u+v}=\frac{1+2\dfrac{v} {u}}{2+\dfrac{v}{u}}$

Now you have a first order homogeneous equation...can you proceed?

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