# Homogeneous or not.

• Oct 26th 2008, 04:43 AM
dE_logics
Homogeneous or not.
To solve non homogeneous differential equations we first have to convert it to homogeneous differential equations (some cases).

So I encountered its problems, or the examples of this type given in my source.

After replacing x + y as u we got the equation -

(2u + 1)du = (3u + 2)dx

In another example (after assuming x + 2y as u)-
(2u + 3)du = (4u + 5)dx

None of the above 2 equations are homogeneous differential equations as claimed in the type's heading, do we have an explanation for this?
• Oct 26th 2008, 06:33 AM
Chris L T521
Quote:

Originally Posted by dE_logics
To solve non homogeneous differential equations we first have to convert it to homogeneous differential equations (some cases).

So I encountered its problems, or the examples of this type given in my source.

After replacing x + y as u we got the equation -

(2u + 1)du = (3u + 2)dx

In another example (after assuming x + 2y as u)-
(2u + 3)du = (4u + 5)dx

None of the above 2 equations are homogeneous differential equations as claimed in the type's heading, do we have an explanation for this?

What were the two original equations?

--Chris
• Oct 26th 2008, 07:21 PM
dE_logics
Equation 1 -

$
\frac{dy}{dx} = \frac{x + y + z}{2x + 2y + 1}
$

Equation 2 -
$
\frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3}
$
• Oct 27th 2008, 02:01 AM
roshanhero
In general I think eqn 2 will be the best one to explain this.
You can change it into homogeneous simply by substituting x=X+h and y=Y+k.
So after differeniating both you would get dx=dX and dy=dY
After substituting it into the eqn you would get
dY/dX=(X+h)+2(Y+k)+1/2(X+h)+4(Y+K)+3
Separate all the X,Y,h and k terms.
After that I hope you would be able to proceed
• Oct 27th 2008, 04:27 AM
dE_logics
aaa...well...actually, these are the procedures followed when
$
\frac{a}{a1} <> \frac{b}{b1}.
$

And you gave the solution, I was asking why (2u + 3)du = (4u + 5)dx is considered homogeneous in my source, I mean they converted
$

$
$\frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3}$
$

" alt="

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to
(2u + 3)du = (4u + 5)dx to make the question homogeneous, which it clearly is not.
• Oct 27th 2008, 09:48 PM
dE_logics
Now I got no idea why <font face="monospace"><em>[
• Oct 29th 2008, 08:07 PM
dE_logics
C'mon people, pls!!
• Oct 29th 2008, 09:04 PM
Chris L T521
Quote:

Originally Posted by dE_logics
Equation 2 -
$
\frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3}
$

Let $u=x+2y\implies\frac{\,du}{\,dx}=1+2\frac{\,dy}{\,d x}\implies \frac{\,dy}{\,dx}=\tfrac{1}{2}\left(\frac{\,du}{\, dx}-1\right)$

Thus, the DE becomes:

$\tfrac{1}{2}\left(\frac{\,du}{\,dx}-1\right)=\frac{u+1}{2u+3}$

Solving for $\frac{\,du}{\,dx}$, we get $\frac{\,du}{\,dx}=\frac{2u+2}{2u+3}+\frac{2u+3}{2u +3}\implies \frac{\,du}{\,dx}=\frac{4u+5}{2u+3}$ $\implies (2u+3)\,du=(4u+5)\,dx\implies (4u+5)\,dx-(2u+3)\,du=0$

From looking at it, its not homogeneous.

To test and see if a DE of this form is homogeneous do this:

Since our DE is in the form of $M(x,u)\,dx+N(x,u)\,du=0$

Evaluate $M(tx,tu)$ and see if you end up with $t^{\alpha}M(x,u)$ and evaluate $N(tx,tu)$ and see if you end up with $t^{\alpha}N(x,u)$. If you can get the powers to be the same, then its homogeneous. If you don't get the same exponents, then its not homogeneous. Its also non-Homogeneous when you can't get $M(tx,tu)=t^{\alpha}M(x,u)$ or $N(tx,tu)=t^{\alpha}N(x,u)$

However, if you need to solve this, you would use separation of variables. Even if you were to solve this, I don't think you will end up with an explicit solution.

--Chris
• Oct 31st 2008, 05:36 AM
dE_logics
Actually I'm having problems in that technique too (Headbang)

Since a^alpha will yield all numbers (accept something opposite to the polarity of a).

So according to this, almost all equations are homogeneous!

This problem was already listed in my long doubt folder.