exact eqn: (e^(x^2/2)*x^2+e^(x^2/2)*x*y^2+e^(x^2/2))dx+....

[rooney]

(e^(x^2/2)*x^2+e^(x^2/2)*x*y^2+e^(x^2/2))dx+2y*e^(x^2/2)=0

I solved for F(x,y) and got y^2*e^(x^2/2)+c

I am now trying to solve for the constant of integration, and found that C'= e^(x^2/2)*x^2+e^(x^2/2)

I don't know how to integrate this properly.

 

[galactus]

Re: exact equation

I am not quite following. Do you want \(\displaystyle \int{x^{2}e^{\frac{x^{2}}{2}}+e^{\frac{x^{2}}{2}}}dx\)?

This is interesting. This is not easily done by elementary means. Yet, the derivative of \(\displaystyle xe^{\frac{x^{2}}{2}}\) is \(\displaystyle (x^{2}+1)e^{\frac{x^{2}}{2}}\).

\(\displaystyle {\therefore}\int{x^{2}e^{\frac{x^{2}}{2}}+e^{\frac{x^{2}}{2}}}dx=xe^{\frac{x^{2}}{2}}\)

So, let's go with that.

 

[rooney]

Re: exact equation

That is the integral I was trying to solve, yes. Thank you very much.

 

[galactus]

Re: exact equation

If you're still interested, here's a little known method that comes in handy. Been awhile snce I used it.

This particular integrand is difficult with parts or substitution. Let's try this.

Whenever we have a polynomial of some sort multiplied by e, we can use the derivative and equate coefficients.

Let \(\displaystyle P(x)=ax^{2}+bx+c\)

\(\displaystyle \frac{d}{dx}[ax^{2}+bx+x]e^{\frac{x^{2}}{2}}=(x^{2}+1)e^{\frac{x^{2}}{2}}\)

Now, use the product rule and get:

\(\displaystyle (2ax+b)e^{\frac{x^{2}}{2}}+(ax^{2}+bx+c)xe^{\frac{x^{2}}{2}}=(x^{2}+1)e^{\frac{x^{2}}{2}}\)

If we equate coefficients, we get:

\(\displaystyle 2a=0, \;\ b+c=1, \;\ b=1, \;\ a=0\)

It's not difficult to see that \(\displaystyle a=0, \;\ b=1, \;\ c=0\)

So, we have:

\(\displaystyle \frac{d}{dx}\left[xe^{\frac{x^{2}}{2}}\right]=(x^{2}+1)e^{\frac{x^{2}}{2}}\)

If we integrate, we get:

\(\displaystyle \boxed{xe^{\frac{x^{2}}{2}}=\int{(x^{2}+1)e^{\frac{x^{2}}{2}}dx}}\)

What is something is I can run this through my calculator and it will not give me an answer for the indefinite integral.

 

[tkhunny]

That's funny. "Undetermined Coefficients" is very common in Differential Equations. I'm not sure I've seen it in an early calculus text. Very nice.

 

[galactus]

Thanks, TKH. I don't know if the poster will appreciate it or not. I think it has a name; I think it's called 'recognition integration' or 'integration by recognition' or something like that. It works well if you have a polynomial multiplied by e. Something like \(\displaystyle \int{P(x)e^{x)}dx\). I think it also works with some trig functions.