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ironictoaster
ODE problem confused!
Confused on b(ii)
I'm not sure what to do.
I have to find u somehow I reckon, not 100% though.
Hmm I believe you just have to show that the identity is valid, because the roots are identical.
pretty similar to the part before it
But I will try first.
the other solution will have an additional x infront of it I am not mistaken.
juts plug it in and see what happens
If it's double roots then yes there's extra x in the solution.
I'm still not sure what should I do first.
I don't know what u is.
assume that \( x(t) = u(t) x_1(t) \) is another solution ... find the value of u(t) ... so that you have complete solution.
@experimentX I genuinely don't know where to start.
i think there is an example in the wikipedia ... in the link i posted above.
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.
|dw:1344360084786:dw|
Note that \[ \Large 4r^2-4r+1=0 \] Has a discriminant of zero.
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 (-:
and how does that prove it equals 0?
Yeah it pretty bad this week
What I understand so far, we need to sub ux into equation 4
\[4\left( u''e^{\frac{t}{2}} + \frac{1}{2}e^{\frac{t}{2}}u' + \frac{1}{4}e^{\frac{t}{2}}u + \frac{1}{2}u'e^{\frac{t}{2}}\right) -4 \left(u' e^{\frac{t}{2}}+ \frac{1}{2}e^{\frac{t}{2}}u \right)+ ue^{\frac{t}{2}}=0\]
Divide by \( \large e^{\frac{t}{2}} \) and then see what happens when you bring it all together.