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

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1. Astrophysics

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 $y''(t)=v''(t)t+v'(t)$

2. Astrophysics

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) $v'=t^{-3}+C_1 \implies v(t) = -\frac{ t^{-2} }{ 2 }+C_2$

3. Astrophysics

But the back of the book says the second solution is $y_2 = t^{-2}$

4. anonymous

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 $y_2 = y_1 \int\limits \frac{ e^{- \int\limits p(t) dt} }{ y_1^2 }dt$

5. Astrophysics

I would do that, but we haven't learnt it haha

6. anonymous

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.

7. Astrophysics

Yeah, we have to use this haha

8. anonymous

Ah, darn. Sorry :(

9. Astrophysics

It's cool! Thanks though!

10. Astrophysics

$y_1(t) = t~~~y_1'(t)=1~~~y_1''(t)=0$ uhh

11. dan815

cauchy form

12. dan815

try t^n solution

13. Astrophysics

No man I can't use that

14. dan815

o ok

15. ShadowLegendX

i lag

16. freckles

$t^2 y''+2t y'-2y=0 \\ t>0 \\ y_1(t)=t \\ \text{ assume } y_2(t)=v \cdot y_1=tv \\ y_2'=tv'+v \\ y_2''=v'+tv''+v'=tv''+2v' \\ \text{ plugging in } \\ t^2(t v''+2 v')+2t(t v'+v)-2(tv)=0 \\ t^3 v''+2t^2v'+2t^2v'+2tv-2tv=0 \\ t^3 v''+4t^2 v'=0 \\ t>0 \\ \text{ so we have } \\ v''+4v'=0$

17. dan815

how come this always happens

18. freckles

you have 2 instead of 4 there

19. dan815

well now thats just a first order ode with a sub

20. freckles

$\text{ oops type-0 } \\ t v''+4v'=0$

21. Astrophysics

Oh you know what I realized my second derivative is wrong as well wow

22. Astrophysics

Ooh ok I see freckles, thanks a million!

23. dan815

like variation of parameter thing if we are plugging in a real solution the lowest order v should always drop out?

24. freckles

$u=v' \\ t u'+4u=0 \\ t \frac{du}{dt}=-4 u \\ \frac{du}{u} =\frac{-4}{ t} dt \\ \ln|u|=-4 \ln|t|+k \\ \ln|u|=-4 \ln|t|+\ln|C| \\ \ln|u|=\ln|t^{-4} C| \\ u=t^{-4} C \\ v'=t^{-4} C$

25. Astrophysics

Yes, that should work :), thanks @freckles

26. 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

27. Astrophysics

Haha, yeah I made it twice actually :x eek, I usually find the mistake when I write it on here as well

28. freckles

actually no I just didn't know how to add 2 and 2 for some reason

29. Astrophysics

xD

30. freckles

I did the differentiating right but for some reason when I got to 2+2 I put 3 down for some reason

31. Astrophysics

Haha silly arithmetic, thanks again!

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