## Astrophysics one year ago @ganeshie8

1. Astrophysics

How do you find f(x) for a second order dif equation to make it exact, I have the equation $y''+xy'+y=0$

2. Astrophysics

I know if you have the form $P(x)y''+Q(x)y'+R(x)y=0$ then the exact equation can be written in the form $[P(x)y']y'+[f(x)y]'=0$

3. Astrophysics

Hmm, I think I can just multiply by an integrating factor

4. Astrophysics

In this case it's $\huge e^{\int\limits Q(x)dx}$ right since it's second order

5. Astrophysics

Haha, nope that's not how it works, I definitely to make it first order

6. Astrophysics

oooh wait...do I see a product rule

7. IrishBoy123

yeah! $$y''+xy'+y=0$$ $$y''+(xy)^{\prime}=0$$ $$y' + xy = C$$ then integrating factor which is $$\exp (\int x \, dx) = e^{{x^2 \over 2}}$$ so $$(y \cdot e^{x^2 \over 2})^{\prime} = C\cdot e^{x^2 \over 2}$$ $$y \cdot e^{x^2 \over 2} = C \int \, e^{x^2 \over 2} \, dx$$ from wiki: $$\text{... there is no elementary indefinite integral for } \int e^{-x^2}\,dx,$$

8. Astrophysics

$(xy)' = xy'+y$

9. Astrophysics

Haha yes! I just figured it out

10. Astrophysics

Thanks @IrishBoy123

11. Astrophysics

Well this is neat xD