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ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0Confused on b(ii)

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what to do.

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0I have to find u somehow I reckon, not 100% though.

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

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0pretty similar to the part before it

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0But I will try first.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0the other solution will have an additional x infront of it I am not mistaken.

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0juts plug it in and see what happens

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0If it's double roots then yes there's extra x in the solution.

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0I'm still not sure what should I do first.

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0I don't know what u is.

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

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX I genuinely don't know where to start.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0i think there is an example in the wikipedia ... in the link i posted above.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0If 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.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1344360084786:dw

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0Pardon 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 (:

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0and how does that prove it equals 0?

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah it pretty bad this week

ironictoaster
 2 years ago
Best ResponseYou've already chosen the best response.0What I understand so far, we need to sub ux into equation 4

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

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