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Astrophysics
 one year ago
@ganeshie8
Astrophysics
 one year ago
@ganeshie8

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1How do you find f(x) for a second order dif equation to make it exact, I have the equation \[y''+xy'+y=0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \[[P(x)y']y'+[f(x)y]'=0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, I think I can just multiply by an integrating factor

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1In this case it's \[\huge e^{\int\limits Q(x)dx}\] right since it's second order

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, nope that's not how it works, I definitely to make it first order

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1oooh wait...do I see a product rule

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3yeah! \(y''+xy'+y=0\) \(y''+(xy)^{\prime}=0\) \(y' + xy = C\) 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: \(\text{... there is no elementary indefinite integral for } \int e^{x^2}\,dx,\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[(xy)' = xy'+y\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha yes! I just figured it out

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Thanks @IrishBoy123

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well this is neat xD
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