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How to solve the differential equation
dy/dx=xy+y^2
In my case this leads to an integral which is unsolvable!
dy/dx=xy+y^2
In my case this leads to an integral which is unsolvable!

Alright, this is how I see it starting from:heman said:That's exactly what i have done,but later it is unsolvable,or rather solution is not clear to me.
Thanks Saltysaltydog 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.
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:heman said:Thanks Salty![]()
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!