- #1

- 2

- 0

Can anyone help with the following:

dy/dx = ay / (bx

a,b constants

thanks,

dy/dx = ay / (bx

^{2}+xy )a,b constants

thanks,

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- #1

- 2

- 0

Can anyone help with the following:

dy/dx = ay / (bx^{2} +xy )

a,b constants

thanks,

dy/dx = ay / (bx

a,b constants

thanks,

- #2

- 800

- 34

- #3

- 1,805

- 54

The key is to consider the unknown function x(y) instead of y(x)

Hi. Thanks for that. However, I'm not convinced your use of the Ei is correct if you define:

[tex]Ei(x)=\int_{-\infty}^x \frac{e^t}{t}dt[/tex]

If I start with the expression:

[tex]d(e^y t)=-\frac{e^y}{y}[/tex]

then integrating from y_0, t_0 to y,t:

[tex]\int_{\substack{t=t_0 \\ y=y_0}}}^{y,t}d(e^y t)=-\int_{y_0}^y \frac{e^y}{y}dy[/tex]

and get:

[tex]x(y)=\frac{e^y}{\frac{e^{y_0}}{x_0}-\int_{y_0}^y \frac{e^y}{y}dy}[/tex]

Now, that agrees with numerical calculations but I'm not sure how I could express that in terms of Ei(x) though.

- #4

- 800

- 34

Of course, you cannot express that in terms of Ei(x), but in terms of Ei(y).that agrees with numerical calculations but I'm not sure how I could express that in terms of Ei(x) though

May be, writing "in terms of" isn't the good wording. What I mean is that Ei(y) is the special function involved in the formula for x(y), as it was shown.

But I never said that Ei(x) is involved in an hypothetical formula for y(x). On the contrary, I said that the analytical inversion of x(y) in order to obtain y(x) is probably utopian with a finite number of elementary functions and even with classical special functions.

- #5

- 2

- 0

thanks guys! - I appreciate the help

- #6

- 1,805

- 54

Of course, you cannot express that in terms of Ei(x), but in terms of Ei(y).

May be, writing "in terms of" isn't the good wording. What I mean is that Ei(y) is the special function involved in the formula for x(y), as it was shown.

But I never said that Ei(x) is involved in an hypothetical formula for y(x). On the contrary, I said that the analytical inversion of x(y) in order to obtain y(x) is probably utopian with a finite number of elementary functions and even with classical special functions.

Ok, I messed that up. I meant Ei(y) but I see as long as you define Ei(y) appropriately, I see what you mean.

Here's the numerical check of:

[tex]y'=\frac{y}{x^2+xy},\quad y(1)=1[/tex]

Code:

```
mysol = NDSolve[{Derivative[1][y][x] ==
y[x]/(x^2 + x*y[x]), y[1] == 1}, y,
{x, 1, 5}];
p1 = Plot[y[x] /. mysol, {x, 1, 5}];
myx[y_] := Exp[y]/(Exp[1] -
NIntegrate[Exp[u]/u, {u, 1, y}]);
mytable = Table[{myx[y], y},
{y, 1, 1.6, 0.01}];
p2 = ListPlot[mytable, Joined -> True];
Show[{p1, p2}]
```

Notice how I tabulate the numbers in terms of {x(y),y} to retrieve the inverse numerically. The two plots superimpose nicely.

- #7

- 800

- 34

For me where was no doubt.

Nevertheless its a good idea to check it numerically. Well donne !

Nevertheless its a good idea to check it numerically. Well donne !

- #8

- 66

- 0

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