MHB 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! :) :) :)