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Homogenous differential equation

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  • #1
etf
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My book says:
"Differential equations of the form $$\frac{\mathrm{d} y}{\mathrm{d} x}=f(x,y)$$, where $$f(x,y)$$ is homogenous function (function is homogenous if $$f(tx,ty)=t^k f(x,y)$$) can be written in form $$\frac{\mathrm{d} y}{\mathrm{d} x}=F(\frac{y}{x})$$ and transformed to differential equation with separate variables using substitution $$\frac{y}{x}=z(x)$$."
Here is my differential equation and how I tried to solve it:
$$(x+y)^2 \frac{\mathrm{d} y}{\mathrm{d} x}=r^2$$
$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}$$
$$f(x,y)=\frac{r^2}{(x+y)^2}$$
$$\rightarrow f(tx,ty)=\frac{r^2}{(tx+ty)^2}=\frac{r^2}{(t(x+y))^2}=\frac{r^2}{t^2(x+y)^2}=t^{-2}\frac{r^2}{(x+y)^2}$$ so our function is homogenous and we can solve our diff. eq., according to my book, using substitution $$\frac{y}{x}=z(x)$$
$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}=\frac{r^2}{(x(1+\frac{y}{x}))^2}=\frac{r^2}{x^2(1+\frac{y}{x})^2}$$
Our equation now becomes $$z+x\frac{\mathrm{d} z}{\mathrm{d} x}=\frac{r^2}{x^2(1+z)^2}$$ but this is not diff. eq. with separate variables? :confused:
 
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  • #2
HallsofIvy
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My book says:
"Differential equations of the form $$\frac{\mathrm{d} y}{\mathrm{d} x}=f(x,y)$$, where $$f(x,y)$$ is homogenous function (function is homogenous if $$f(tx,ty)=t^k f(x,y)$$) can be written in form $$\frac{\mathrm{d} y}{\mathrm{d} x}=F(\frac{y}{x})$$ and transformed to differential equation with separate variables using substitution $$\frac{y}{x}=z(x)$$."
This is incorrect. A differential equation is "homogeneous of order k" if [itex]f(tx, ty)= t^k f(x, y)[/itex]. A differential equation that is homogenous of order 0 can be written as a separable equation in the variable z= y/x.

Here is my differential equation and how I tried to solve it:
$$(x+y)^2 \frac{\mathrm{d} y}{\mathrm{d} x}=r^2$$
$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}$$
$$f(x,y)=\frac{r^2}{(x+y)^2}$$
$$\rightarrow f(tx,ty)=\frac{r^2}{(tx+ty)^2}=\frac{r^2}{(t(x+y))^2}=\frac{r^2}{t^2(x+y)^2}=t^{-2}\frac{r^2}{(x+y)^2}$$ so our function is homogenous
No. "Homogeneous", with no adjective, means "homogeneous of degree 0". That is, f(tx, ty)= f(x, y) for any t. That is the condition that we can write the problem as a differential equation in z= y/x. Your equation is homogeneous of order -2 so cannot be written that way.

and we can solve our diff. eq., according to my book, using substitution $$\frac{y}{x}=z(x)$$
$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}=\frac{r^2}{(x(1+\frac{y}{x}))^2}=\frac{r^2}{x^2(1+\frac{y}{x})^2}$$
Our equation now becomes $$z+x\frac{\mathrm{d} z}{\mathrm{d} x}=\frac{r^2}{x^2(1+z)^2}$$ but this is not diff. eq. with separate variables? :confused:
 
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  • #3
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Well, I am not that familiar with the theory, but you say yourself that you need to write your equation on the form
\begin{equation*}
\frac{dy}{dx} = F(\frac{y}{x}).
\end{equation*}
Does this seem doable?

That said, my first instinct would instead be to try the substitution ##u = x + y##.
 
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  • #4
LCKurtz
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My book says:
"Differential equations of the form $$\frac{\mathrm{d} y}{\mathrm{d} x}=f(x,y)$$, where $$f(x,y)$$ is homogenous function (function is homogenous if $$f(tx,ty)=t^k f(x,y)$$) can be written in form $$\frac{\mathrm{d} y}{\mathrm{d} x}=F(\frac{y}{x})$$ and transformed to differential equation with separate variables using substitution $$\frac{y}{x}=z(x)$$."
Here is my differential equation and how I tried to solve it:
$$(x+y)^2 \frac{\mathrm{d} y}{\mathrm{d} x}=r^2$$
And what is ##r^2##? Just any constant or does it happen to be ##x^2+y^2##? That would make your DE homogeneous.
 
  • #5
etf
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Thanks for replies. I solved it using substitution x+y=z(x). Btw r^2 is constant.
 

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