# Thread: First order ODE getting to homogenous with x and y substitutions

1. ## First order ODE getting to homogenous with x and y substitutions

Hi folks,

by substituting $x = X + 1$ and $y = Y + 3$ reduce the differential equation

$\dfrac{dy}{dx} = \dfrac{4x - y + 7}{2x + y - 1}$ ................... (1)

to a homogeneous equation and hence find the general solution in terms of x and y.

Substituting the values of x and y into (1) we get

$\dfrac{dy}{dx} = \dfrac{4X - Y}{2X + Y}$ which can be made homogeneous, that is, a function of $\frac{Y}{X}$ by dividing top and bottom by X.

Once we have a homogeneous equation, it can be solved using the y = vx substitution, but first I have to get $\frac{dy}{dx}$ into the form $\frac{DY}{DX}$. I cannot see how to do this. I have tried different combinations of the chain rule, but not getting anywhere.

A hint would really help!

Thanks

2. ## Re: First order ODE getting to homogenous with x and y substitutions

Your original equation is $\displaystyle \frac{dy}{dx}= \frac{4x- y+ 7}{2x+ y- 1}$. You make the substitution x= X- 1 (NOT "x= X+ 1"), y= Y+ 3. $\displaystyle \frac{dy}{dx}= \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}= \frac{dY}{dX}$. 4x-y+ 7= 4(X- 1)- Y- 3+ 7= 4X- Y and 2x+ y- 1= 2(X- 1)+ Y+ 3- 1= 2X+ Y. Yes, that makes the equation $\displaystyle \frac{dY}{dX}= \frac{4X- Y}{2X+ Y}$.

As to how I went from $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{dY}{dx}$, first I replaced y by Y+ 3: $\displaystyle \frac{dY}{dx}= \frac{Y+ 3}{dx}= \frac{dY}{dx}+ 0= \frac{dY}{dx}$. Now to go from $\displaystyle \frac{dY}{dx}$ to $\displaystyle \frac{dY}{dX}$, use the chain rule: $\displaystyle \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}$ and $\displaystyle \frac{dX}{dx}= 1$ because X= x- 1.

Dividing both numerator and denominator of that fraction by X gives $\displaystyle \frac{dY}{dX}= \frac{4-\frac{Y}{X}}{2+ \frac{Y}{X}}$. Let V= Y/X. Then Y= XV and $\displaystyle \frac{dY}{dX}= V+ X\frac{dV}{dX}= \frac{4- V}{2+ V}$. Then $\displaystyle \frac{2+ V}{4- V}\frac{dV}{dX}= dX$.

First do the "division" indicated by the fraction. 4- V divides into 2+ V negative one times with remainder -2. That is, $\displaystyle \frac{2+ V}{4- V}= -1- \frac{2}{4- V}$. Integrate that with respect to V.

3. ## Re: First order ODE getting to homogenous with x and y substitutions

Excellent! Many thanks, HallsofIvy, once again! I made a typo, you are right the problem said x = X - 1.

4. ## Re: First order ODE getting to homogenous with x and y substitutions

I think you made a mistake at the end.

$X \dfrac{dV}{dX} = \dfrac{4 - V}{2 + V} - V$ which gives $X \dfrac{dV}{dX} = \dfrac{(4 + V)(1 - V)}{(2 + V)}$ but this is just mechanics. Many thanks for setting me on the right path.

5. ## Re: First order ODE getting to homogenous with x and y substitutions

Originally Posted by HallsofIvy
Your original equation is $\displaystyle \frac{dy}{dx}= \frac{4x- y+ 7}{2x+ y- 1}$. You make the substitution x= X- 1 (NOT "x= X+ 1"), y= Y+ 3. $\displaystyle \frac{dy}{dx}= \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}= \frac{dY}{dX}$. 4x-y+ 7= 4(X- 1)- Y- 3+ 7= 4X- Y and 2x+ y- 1= 2(X- 1)+ Y+ 3- 1= 2X+ Y. Yes, that makes the equation $\displaystyle \frac{dY}{dX}= \frac{4X- Y}{2X+ Y}$.

As to how I went from $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{dY}{dx}$, first I replaced y by Y+ 3: $\displaystyle \frac{dY}{dx}= \frac{Y+ 3}{dx}= \frac{dY}{dx}+ 0= \frac{dY}{dx}$. Now to go from $\displaystyle \frac{dY}{dx}$ to $\displaystyle \frac{dY}{dX}$, use the chain rule: $\displaystyle \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}$ and $\displaystyle \frac{dX}{dx}= 1$ because X= x- 1.

Dividing both numerator and denominator of that fraction by X gives $\displaystyle \frac{dY}{dX}= \frac{4-\frac{Y}{X}}{2+ \frac{Y}{X}}$. Let V= Y/X. Then Y= XV and $\displaystyle \frac{dY}{dX}= V+ X\frac{dV}{dX}= \frac{4- V}{2+ V}$. Then $\displaystyle \frac{2+ V}{4- V}\frac{dV}{dX}= dX$.
Oh, bother! I did, in fact, forget the "V" term on the left: $\displaystyle X\frac{dV}{dX}= \frac{4- V}{2+ V}- V= \frac{4- V}{2+ V}- \frac{2V+ V^2}{2+ V}= \frac{4- 3V- V^2}{2+ V}= \frac{(4- V)(1+ V)}{2+ V}$.

$\displaystyle \frac{2+ V}{(4- V)(1+ V)}dV= \frac{1}{X}dX$.

First do the "division" indicated by the fraction. 4- V divides into 2+ V negative one times with remainder -2. That is, $\displaystyle \frac{2+ V}{4- V}= -1- \frac{2}{4- V}$. Integrate that with respect to V.

6. ## Re: First order ODE getting to homogenous with x and y substitutions

I get

$\dfrac{4 - 3V - V^2}{2 + V} = \dfrac{(4 + V)(1 - V)}{2 + V}$