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Astrophysics
 one year ago
Reduction of order (To find second solution)
Astrophysics
 one year ago
Reduction of order (To find second solution)

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I have to use the method of reduction order to find a second solution to this equation \[t^2y''+2ty'2y=0, ~~~t>0;y_1(t)=t\] I checked this solution it checks out but the one I got...\[y(t)=v(t)y_1(t)=v(t)t\] \[y'(t)=v'(t)t+v(t)\] and \[y''(t)=v''(t)t+v'(t)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I set the equation up as \[y'' + \frac{ 2 }{ t }y'\frac{ 2 }{ t^2 }y=0\] and then I plugged in above \[v''(t)t+v'(t)+\frac{ 2 }{ t } \left[ v'(t)t+v(t) \right]\frac{ 2 }{ t^2 }v(t)t=0 \implies v''(t)t+2v'(t)=0\] then I made the substitutions v'(t)=u(t), and v''(t)=u'(t) making my equation \[u'(t)t+3u(t)=0 \implies u't+3u=0\]\[u=t^{3}+C_1~~~~C_1=e^C\] now subbing back v'(t)=u(t) \[v'=t^{3}+C_1 \implies v(t) = \frac{ t^{2} }{ 2 }+C_2\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1But the back of the book says the second solution is \[y_2 = t^{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I actually figured out a bit of a simpler way to solve for a secondary solution for second order differential equation with nonconstant coefficients. See here: http://openstudy.com/study#/updates/561b2901e4b07ab19da73807 The professor showed using \[y_2 = y_1 \int\limits \frac{ e^{ \int\limits p(t) dt} }{ y_1^2 }dt\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I would do that, but we haven't learnt it haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does your professor want you to strictly use the method they taught? If not, this might save time. =/ I haven't learned how to solve with nonconstant coefficients yet, so I can't exactly help. But from the other thread I learned how to solve using that method.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, we have to use this haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1It's cool! Thanks though!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[y_1(t) = t~~~y_1'(t)=1~~~y_1''(t)=0\] uhh

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1No man I can't use that

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[t^2 y''+2t y'2y=0 \\ t>0 \\ y_1(t)=t \\ \text{ assume } y_2(t)=v \cdot y_1=tv \\ y_2'=tv'+v \\ y_2''=v'+tv''+v'=tv''+2v' \\ \text{ plugging in } \\ t^2(t v''+2 v')+2t(t v'+v)2(tv)=0 \\ t^3 v''+2t^2v'+2t^2v'+2tv2tv=0 \\ t^3 v''+4t^2 v'=0 \\ t>0 \\ \text{ so we have } \\ v''+4v'=0 \]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0how come this always happens

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you have 2 instead of 4 there

dan815
 one year ago
Best ResponseYou've already chosen the best response.0well now thats just a first order ode with a sub

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\text{ oops type0 } \\ t v''+4v'=0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh you know what I realized my second derivative is wrong as well wow

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ooh ok I see freckles, thanks a million!

dan815
 one year ago
Best ResponseYou've already chosen the best response.0like variation of parameter thing if we are plugging in a real solution the lowest order v should always drop out?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[u=v' \\ t u'+4u=0 \\ t \frac{du}{dt}=4 u \\ \frac{du}{u} =\frac{4}{ t} dt \\ \lnu=4 \lnt+k \\ \lnu=4 \lnt+\lnC \\ \lnu=\lnt^{4} C \\ u=t^{4} C \\ v'=t^{4} C\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yes, that should work :), thanks @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.4np lol for some reason when I did it on paper I made the same mistake you did but I was like whatever made I made a mistake so I just started over typing here

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, yeah I made it twice actually :x eek, I usually find the mistake when I write it on here as well

freckles
 one year ago
Best ResponseYou've already chosen the best response.4actually no I just didn't know how to add 2 and 2 for some reason

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I did the differentiating right but for some reason when I got to 2+2 I put 3 down for some reason

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha silly arithmetic, thanks again!
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