Reduction of order (To find second solution)

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Reduction of order (To find second solution)

Mathematics
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I 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 \[y''(t)=v''(t)t+v'(t)\]
I 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) \[v'=t^{-3}+C_1 \implies v(t) = -\frac{ t^{-2} }{ 2 }+C_2\]
But the back of the book says the second solution is \[y_2 = t^{-2}\]

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I 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 \[y_2 = y_1 \int\limits \frac{ e^{- \int\limits p(t) dt} }{ y_1^2 }dt\]
I would do that, but we haven't learnt it haha
Does 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.
Yeah, we have to use this haha
Ah, darn. Sorry :(
It's cool! Thanks though!
\[y_1(t) = t~~~y_1'(t)=1~~~y_1''(t)=0\] uhh
cauchy form
try t^n solution
No man I can't use that
o ok
i lag
\[t^2 y''+2t y'-2y=0 \\ t>0 \\ y_1(t)=t \\ \text{ assume } y_2(t)=v \cdot y_1=tv \\ y_2'=tv'+v \\ y_2''=v'+tv''+v'=tv''+2v' \\ \text{ plugging in } \\ t^2(t v''+2 v')+2t(t v'+v)-2(tv)=0 \\ t^3 v''+2t^2v'+2t^2v'+2tv-2tv=0 \\ t^3 v''+4t^2 v'=0 \\ t>0 \\ \text{ so we have } \\ v''+4v'=0 \]
how come this always happens
you have 2 instead of 4 there
well now thats just a first order ode with a sub
\[\text{ oops type-0 } \\ t v''+4v'=0\]
Oh you know what I realized my second derivative is wrong as well wow
Ooh ok I see freckles, thanks a million!
like variation of parameter thing if we are plugging in a real solution the lowest order v should always drop out?
\[u=v' \\ t u'+4u=0 \\ t \frac{du}{dt}=-4 u \\ \frac{du}{u} =\frac{-4}{ t} dt \\ \ln|u|=-4 \ln|t|+k \\ \ln|u|=-4 \ln|t|+\ln|C| \\ \ln|u|=\ln|t^{-4} C| \\ u=t^{-4} C \\ v'=t^{-4} C\]
Yes, that should work :), thanks @freckles
np 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
Haha, yeah I made it twice actually :x eek, I usually find the mistake when I write it on here as well
actually no I just didn't know how to add 2 and 2 for some reason
xD
I did the differentiating right but for some reason when I got to 2+2 I put 3 down for some reason
Haha silly arithmetic, thanks again!

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