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ironictoasterBest ResponseYou've already chosen the best response.0
Confused on b(ii)
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
I'm not sure what to do.
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
I have to find u somehow I reckon, not 100% though.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Hmm I believe you just have to show that the identity is valid, because the roots are identical.
 one year ago

colorfulBest ResponseYou've already chosen the best response.0
pretty similar to the part before it
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
But I will try first.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
the other solution will have an additional x infront of it I am not mistaken.
 one year ago

colorfulBest ResponseYou've already chosen the best response.0
juts plug it in and see what happens
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Reduction_of_order
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
If it's double roots then yes there's extra x in the solution.
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
I'm still not sure what should I do first.
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
I don't know what u is.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
assume that \( x(t) = u(t) x_1(t) \) is another solution ... find the value of u(t) ... so that you have complete solution.
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
@experimentX I genuinely don't know where to start.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i think there is an example in the wikipedia ... in the link i posted above.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
If I understand this problem then they just want you to check what happens if you substitute back their provided result. I believe their are trying to introduce you to the method of Reduction of Order You will get a result in the form of \[ \Large u(x)=d_1x+d_2 \] where \(d_1, d_2\) are constant. The second solution is of the form \[ \Large y_2(x)=u(x)e^{\frac{x}{2}}\] So you can use superposition to get the general solution.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1344360084786:dw
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Note that \[ \Large 4r^24r+1=0 \] Has a discriminant of zero.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Pardon me if I was interrupting something in here, OpenStudy lags horribly for me today so I hit the post button before it crashes me again (:
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
and how does that prove it equals 0?
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
Yeah it pretty bad this week
 one year ago

ironictoasterBest ResponseYou've already chosen the best response.0
What I understand so far, we need to sub ux into equation 4
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
\[4\left( u''e^{\frac{t}{2}} + \frac{1}{2}e^{\frac{t}{2}}u' + \frac{1}{4}e^{\frac{t}{2}}u + \frac{1}{2}u'e^{\frac{t}{2}}\right) 4 \left(u' e^{\frac{t}{2}}+ \frac{1}{2}e^{\frac{t}{2}}u \right)+ ue^{\frac{t}{2}}=0\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Divide by \( \large e^{\frac{t}{2}} \) and then see what happens when you bring it all together.
 one year ago
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