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dy/dx=xy+y^2

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

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- Thread starter heman
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- #1

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dy/dx=xy+y^2

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

- #2

arildno

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It might be there exists a trick to manage this one, but off-hand, I can't help you.

- #3

Zurtex

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- #4

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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

GCT

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

[tex]v=y^{(1-n)}[/tex]

[tex]v=y^{(1-n)}[/tex]

- #6

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

- #7

saltydog

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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

saltydog

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- #9

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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

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

saltydog

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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:

[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.

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- #11

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Thanks Salty!

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