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

## Main Question or Discussion Point

How to solve the differential equation
dy/dx=xy+y^2

In my case this leads to an integral which is unsolvable!

## Answers and Replies

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arildno
Homework Helper
Gold Member
Dearly Missed
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.

Zurtex
Homework Helper
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.

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!

GCT
Homework Helper
It's a Bernoulli equation, you can solve it by Leibniz substitution

$$v=y^{(1-n)}$$

That's exactly what i have done,but later it is unsolvable,or rather solution is not clear to me.

saltydog
Homework Helper
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:

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

Integrating:

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

Yielding:

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

Noting that:

$$\int_{x_0}^x f(t)dt=K+\int_0^x f(t)dt$$

We can write the above expression as:

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

Letting $u=t/\sqrt{2}$ we obtain:

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

Now, it just so happens that:

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

and:

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

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

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

or:

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

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

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

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:

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

would you? Same dif.

saltydog
Homework Helper
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.

saltydog said:
Alright, this is how I see it starting from:

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

Integrating:

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

Yielding:

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

Noting that:

$$\int_{x_0}^x f(t)dt=K+\int_0^x f(t)dt$$

We can write the above expression as:

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

Letting $u=t/\sqrt{2}$ we obtain:

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

Now, it just so happens that:

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

and:

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

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

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

or:

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

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

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

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:

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

would you? Same dif.
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!

saltydog
Homework Helper
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!
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:

$$2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt$$

I can call it Sal[x] such that:

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

and further:

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

Same dif as Sin[x] as far as I'm concerned.

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Thanks Salty!