Reduce an equation to a homogenous equation?

[Cadbury]

Hi, please help me solve this problem :)

Reduce (2x + 3y - 5) dx + (3x - y - 2) dy = 0 into a homogenous equation

I know that a differential equation Mdx + Ndy is homogeneous in x and y and if M and N are homogeneous functions of the same degree but given this, I have no idea where to start. Should I multiply both sides to something? Any help will be appreciated :)
Thank you so much!

 

[Euge]

Hi Cadbury,

I misunderstood your question at first but you can reduce your equation to a homogeneous type by a linear transformation of the form $u = x + A$, $v = y + B$ for constants $A$ and $B$.

 

[chisigma]

Hi, please help me solve this problem :)

Reduce (2x + 3y - 5) dx + (3x - y - 2) dy = 0 into a homogenous equation

I know that a differential equation Mdx + Ndy is homogeneous in x and y and if M and N are homogeneous functions of the same degree but given this, I have no idea where to start. Should I multiply both sides to something? Any help will be appreciated :)
Thank you so much!

Welcome on MHB Cadbury!...

... the procedure is tedious but not difficult!... setting $\displaystyle 2\ x + 3\ y - 5 = u$ and $\displaystyle 3\ x - y - 2 = v$ You obtain first...

$\displaystyle d u = 2\ d x + 3\ d y$

$\displaystyle d v = 3\ d x - d y\ (1)$

Solving (1) ...

$\displaystyle d x = \frac{d u + 3\ d v}{11}$

$\displaystyle d y = \frac{3\ d u - 2\ d v}{11}\ (2)$

... and now from (2) and the original equation...

$\displaystyle \frac{d y}{d x} = \frac{3\ du - 2\ d v}{d u + 3\ d v} = - \frac{u}{v}\ (3)$

... and finally with some steps...

$\displaystyle \frac{d u}{d v} = \frac{2\ v - 3\ u}{3\ v + u}\ (4)$

Kind regards

$\chi$ $\sigma$

 

[Cadbury]

Welcome on MHB Cadbury!...

... the procedure is tedious but not difficult!... setting $\displaystyle 2\ x + 3\ y - 5 = u$ and $\displaystyle 3\ x - y - 2 = v$ You obtain first...

$\displaystyle d u = 2\ d x + 3\ d y$

$\displaystyle d v = 3\ d x - d y\ (1)$

Solving (1) ...

$\displaystyle d x = \frac{d u + 3\ d v}{11}$

$\displaystyle d y = \frac{3\ d u - 2\ d v}{11}\ (2)$

... and now from (2) and the original equation...

$\displaystyle \frac{d y}{d x} = \frac{3\ du - 2\ d v}{d u + 3\ d v} = - \frac{u}{v}\ (3)$

... and finally with some steps...

$\displaystyle \frac{d u}{d v} = \frac{2\ v - 3\ u}{3\ v + u}\ (4)$

Kind regards

$\chi$ $\sigma$

Thank you very much! :) :) :)

 

[blue_raver22]

To reduce an equation to a homogenous equation, we need to make both the coefficients of dx and dy have the same degree. In other words, we want to make the equation have the same degree of homogeneity. To do this, we can divide both sides of the equation by the highest power of x and y present in the equation. In this case, we can divide by (2x + 3y) as it is the highest power present in the equation. This will give us:

dx + [(3x - y - 2)/(2x + 3y)]dy = 0

Now, we can define a new variable u = x/y, which means that x = uy. Substituting this into the equation, we get:

dx + [(3uy - y - 2)/(2uy + 3y)]dy = 0

Factoring out y from the numerator, we get:

dx + [y(3u - 1)/(y)(2u + 3)]dy = 0

Simplifying, we get:

dx + [(3u - 1)/(2u + 3)]dy = 0

This is now a homogenous equation as both the coefficients of dx and dy have the same degree of homogeneity (1).