Question

@ganeshie8

Answer 1

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

Answer 2

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 \[[P(x)y']y'+[f(x)y]'=0\]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

oooh wait...do I see a product rule

Answer 7

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:
\(\text{... there is no elementary indefinite integral for }
\int e^{-x^2}\,dx,\)

Answer 8

\[(xy)' = xy'+y\]

Answer 9

Haha yes! I just figured it out

Answer 10

Thanks @IrishBoy123

Answer 11

Well this is neat xD