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I have started learning ODE and I couldn't solve this exercise: $$\frac{dy}{dx}=\frac{x+y+4}{x-y-6}$$

I am a beginner and only learned separation of variables and how to solve homogeneous equations by using $y=ux$ but I don't think this problem can be solved by any of these.

I tried to use sub $x-y=u$ but I didn't get anything useful from this,

Integreek
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One
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1 Answers1

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Substitute $x=X+h$ and $y=Y+k$ where $h, k$ are constants$\big{(}$, i.e., $\mathrm dx=\mathrm dX$ and $\mathrm dy=\mathrm dY\implies\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dY}{\mathrm dX}\big{)}$ such that the ODE becomes homogeneous:

$$\frac{\mathrm dY}{\mathrm dX}=\frac{X+Y+(h+k+4)}{X-Y+(h-k-6)}$$

For the ODE to be homogeneous, we get the system of equations:

$$h+k=-4$$ $$h-k=6$$

Hence, $h=1, k=-5$.

Now, the ODE becomes:

$$\frac{\mathrm dY}{\mathrm dX}=\frac{X+Y}{X-Y}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$$ Where $X=x-1, Y=y+5$. This can be easily solved by the substitution $Y=tX$.

Integreek
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  • If $x=X+h, y=Y+k$ how to find $\frac{dy}{dx}$ in terms of $\frac{dY}{dX}$? – One Nov 06 '24 at 19:22
  • More importantly: If $x= f(X,Y), y=g(X,Y)$ What is $\frac{dy}{dx}$ in terms of $\frac{dY}{dX}$? – One Nov 06 '24 at 19:31
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    @One In general you need a Jacobian determinant, but in this case $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}Y}{\mathrm{d}X}$. – J.G. Nov 06 '24 at 19:36
  • @J.G. I comepletly forgot about this, I need to revise this any recommendation on what book to use? It would be better that it has a lot of exercises – One Nov 06 '24 at 19:44
  • @One $\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dY}{\mathrm dX}$ since $h$ and $k$ are constants, see my edited answer. – Integreek Nov 07 '24 at 03:10
  • @MathGuy I don't understand what is $dY$ and $dy$ without a fraction (what did you differentiate both of them ?), it is seem like cheating. – One Nov 07 '24 at 13:26
  • @One these are differentials and such notation is absolutely valid. Think of $\mathrm dx$ as an infinitesimal change in $x$. You must've encountered these in chain rule and separation of variables too(while cross-multiplying $\mathrm dx$). – Integreek Nov 07 '24 at 13:33
  • I know what they are but I on't understand how does that works, Is there a rigorous way to prove this? – One Nov 07 '24 at 14:44
  • @One well this is indeed rigorous. $Y$ can be considered as a function of $y$: $Y(y)=y+5\implies \mathrm dY=1\cdot\mathrm dy$ by the definition of differentials. Similarly, it can be shown that $\mathrm dX=\mathrm dx$. – Integreek Nov 07 '24 at 14:50
  • @MathGuy hmmm I think I need to revise, but I don't recall learn about this or using this in real analysis/ calculus – One Nov 07 '24 at 14:56
  • @One also, the concept of differentials is applied to solve exact DEs where exact differentials such as $d(xy)=ydx+xdy$ are widely used. – Integreek Nov 07 '24 at 15:02
  • I don't like this since it appear to be an abuse of notation – One Nov 07 '24 at 15:15
  • @One you may like to see this. – Integreek Nov 07 '24 at 15:26
  • According to me, there is no harm in writing and doing stuff like this unless we try to delve into its philosophical aspects. – Integreek Nov 07 '24 at 15:35
  • I agree that it appears to be an abuse of notation. I've always thought of differentials intuitively as denoting very small changes in variables. I'm in high school, and till now such an understanding seems plausible to me, although I've seen on other threads here on MSE that this is not strictly correct, but for now, even my teacher has said to go by this intuitive definition only. TBH, I'm kinda confused about dx too, but when I go by its intuitive definition, all seems well. – Integreek Nov 07 '24 at 15:41
  • @One you may want to go through these threads: https://math.stackexchange.com/questions/2142783/separable-differential-equations-detaching-dy-dx/2142793#2142793, https://math.stackexchange.com/questions/27425/what-am-i-doing-when-i-separate-the-variables-of-a-differential-equation – Integreek Nov 10 '24 at 04:14
  • @One Did you finally understand how $\mathrm dY=\mathrm dy$ and $\mathrm dX=\mathrm dx$? If that’s still not clear, let me remind you that you would’ve used differentials a lot in $u$-substitutions in integration, eg: $u=\sin x\implies\mathrm du=\cos x\mathrm dx$. – Integreek Dec 07 '24 at 18:33