Solve the differential equation dy/dx=xy+y^2

In summary, the conversation discusses the unsolvability of the differential equation dy/dx=xy+y^2 and proposes different methods to solve it. One person suggests using the Erfi function, while another suggests using a Leibniz substitution. The conversation also touches on the idea of "equal rights for special functions" and the emergence of new functions in mathematics.
  • #1
heman
361
0
How to solve the differential equation
dy/dx=xy+y^2

In my case this leads to an integral which is unsolvable! :cry:
 
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  • #2
Most ODE's are analytically unsolvable.
It might be there exists a trick to manage this one, but off-hand, I can't help you.
 
  • #3
The answer can not be expressed in terms of elementary functions. Although with the use of the quite commonly used Erfi function there is a fairly simple representation of the answer, why don't you show us what you have done so far.
 
  • #4
I have done it this way:
i have put -1/y=t =>1/y^2dy=dt

dt/dx + xt =1
Integrating factor comes out to be e^(x^2/2)
So finally i have to integrate e^(x^2/2) to get the solution.!
I can do the integral but what perplexes is me that it's solution is not in form of elementary functions!
 
  • #5
It's a Bernoulli equation, you can solve it by Leibniz substitution

[tex]v=y^{(1-n)}[/tex]
 
  • #6
That's exactly what i have done,but later it is unsolvable,or rather solution is not clear to me.
 
  • #7
heman said:
That's exactly what i have done,but later it is unsolvable,or rather solution is not clear to me.

Alright, this is how I see it starting from:

[tex]d\left[e^{x^2/2}z\right]=-e^{x^2/2}[/tex]

Integrating:

[tex]\int_{x_0,z_0}^{x,z} d\left[e^{x^2/2}z\right]=-\int_{x_0}^x e^{x^2/2}dx[/tex]

Yielding:

[tex]z=Ke^{-x^2/2}-e^{-x^2/2}\int_{x_0}^x e^{t^2/2}dt[/tex]

Noting that:


[tex]\int_{x_0}^x f(t)dt=K+\int_0^x f(t)dt[/tex]

We can write the above expression as:

[tex]z(y)=Ce^{-x^2/2}-e^{-x^2/2}\int_0^x e^{t^2/2}dt[/tex]

Letting [itex]u=t/\sqrt{2}[/itex] we obtain:

[tex]z(y)=Ce^{-x^2/2}-\sqrt{2}e^{-x^2/2}\int_0^{x/\sqrt{2}} e^{u^2}du\quad\tag{1}[/tex]

Now, it just so happens that:

[tex]Erfi[x]=\frac{Erf[ix]}{i}[/tex]

and:

[tex]Erf[ix]=\frac{2i}{\sqrt{\pi}}\int_0^{x} e^{t^2}dt[/tex]

so the i's cancel and we're left with:

[tex]Erfi[x]=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{t^2}dt[/tex]

or:

[tex]\int_0^{x} e^{t^2}dt=\frac{\sqrt{\pi}}{2}Erfi[x][/tex]

Substituting that into (1) we get what Mathematica reports:

[tex]z(y)=Ce^{-x^2/2}-e^{-x^2/2}\sqrt{\frac{\pi}{2}}
Erfi\left[\frac{x}{\sqrt{2}}\right][/tex]

But z is 1/y so then 1 over all that stuf is the answer (if I didn't make any mistakes). How about a plot?

Also, I'll quote someone in here:

"Equal rights for special functions"

That means, treat Erfi[x] in the same way as Sin[x]. You wouldn't mind if the answer was:

[tex]z(x)=Ce^{-x^2/2}Sin[x][/tex]

would you? Same dif.
 
  • #8
Oh yea, back-substitute everything into the ODE in Mathematica to make sure the answer is right. We can do that . . . like Gauss and the rest of them wouldn't have used Mathematica if they had it back then. Be for real. :smile:
 
  • #9
saltydog said:
Alright, this is how I see it starting from:

[tex]d\left[e^{x^2/2}z\right]=-e^{x^2/2}[/tex]

Integrating:

[tex]\int_{x_0,z_0}^{x,z} d\left[e^{x^2/2}z\right]=-\int_{x_0}^x e^{x^2/2}dx[/tex]

Yielding:

[tex]z=Ke^{-x^2/2}-e^{-x^2/2}\int_{x_0}^x e^{t^2/2}dt[/tex]

Noting that:


[tex]\int_{x_0}^x f(t)dt=K+\int_0^x f(t)dt[/tex]

We can write the above expression as:

[tex]z(y)=Ce^{-x^2/2}-e^{-x^2/2}\int_0^x e^{t^2/2}dt[/tex]

Letting [itex]u=t/\sqrt{2}[/itex] we obtain:

[tex]z(y)=Ce^{-x^2/2}-\sqrt{2}e^{-x^2/2}\int_0^{x/\sqrt{2}} e^{u^2}du\quad\tag{1}[/tex]

Now, it just so happens that:

[tex]Erfi[x]=\frac{Erf[ix]}{i}[/tex]

and:

[tex]Erf[ix]=\frac{2i}{\sqrt{\pi}}\int_0^{x} e^{t^2}dt[/tex]

so the i's cancel and we're left with:

[tex]Erfi[x]=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{t^2}dt[/tex]

or:

[tex]\int_0^{x} e^{t^2}dt=\frac{\sqrt{\pi}}{2}Erfi[x][/tex]

Substituting that into (1) we get what Mathematica reports:

[tex]z(y)=Ce^{-x^2/2}-e^{-x^2/2}\sqrt{\frac{\pi}{2}}
Erfi\left[\frac{x}{\sqrt{2}}\right][/tex]

But z is 1/y so then 1 over all that stuf is the answer (if I didn't make any mistakes). How about a plot?

Also, I'll quote someone in here:

"Equal rights for special functions"

That means, treat Erfi[x] in the same way as Sin[x]. You wouldn't mind if the answer was:

[tex]z(x)=Ce^{-x^2/2}Sin[x][/tex]

would you? Same dif.

Thanks Salty :smile:
Everything is fine and well for me in yours solution!
Could you please throw some light on this New Emergent function!
I came across this first time!
 
  • #10
heman said:
Thanks Salty :smile:
Everything is fine and well for me in yours solution!
Could you please throw some light on this New Emergent function!
I came across this first time!

Well Erfi[x] is called the imaginary Error function and Erf[x] is the error function. Check out Mathworld for details. But just any valid expression can hold for a new function. There's tons of them. For example, if I have some irreducible integral of the form:

[tex]2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]

I can call it Sal[x] such that:

[tex]Sal[x]=2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]

and further:

[tex]\frac{dSal[x]}{dx}=\frac{1}{\sqrt{ln(x)+c}}[/tex]


Same dif as Sin[x] as far as I'm concerned.
 
Last edited:
  • #11
Thanks Salty! :smile:
 

FAQ: Solve the differential equation dy/dx=xy+y^2

1. What is a differential equation?

A differential equation is a mathematical equation that relates the rate of change of a function to the function itself. It is used to model many natural phenomena and is an important tool in science and engineering.

2. How do you solve a differential equation?

To solve a differential equation, you need to find a function that satisfies the equation. This can be done through various methods such as separation of variables, substitution, or using an integrating factor.

3. What does the notation dy/dx mean?

The notation dy/dx represents the derivative of the function y with respect to x. It represents the rate of change of y with respect to x.

4. How do you solve the differential equation dy/dx=xy+y^2?

This differential equation can be solved using the method of separation of variables. First, we can rearrange the equation to get dy/dx=y(x+y). Then, we can separate the variables to get 1/y dy = (x+y)dx. By integrating both sides, we get ln|y|=x^2/2 + xy + C. Finally, we can solve for y to get y(x)=Ce^(x^2/2+xy).

5. What are the applications of solving differential equations?

Differential equations are used to model many real-world phenomena in fields such as physics, chemistry, biology, engineering, and economics. They can be used to predict the behavior of systems over time and are essential in understanding complex systems.

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