D^2 (x) - 3 dx/dt + 2x = 0
D^2 represents second order differential ie d/dt * d/dt
Given that when t=0 , x=0 , dx/dt = 0
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What ideas have you had so far?
Can i do it by this method
Auxiliary equation is given by m^2 - m + 2 = 0
we will get imaginary and distinct value for m...
Right in general. However, from your DE, you should have m^2 - 3m + 2 = 0. This is all within the framework of assuming a solution of the form x = A e^(m t), right?
Iam sorry , i make stupid mistakes. Now you know
I got the C.F as
x = c1 e^2t + c2 e^t
Now what ?
i got m as 1 and 2 , which are distinct and real
so by the formula given by our teacher i got the the above result
Excellent. All you have to do now is find out what the constants c1 and c2 are. You do that by plugging in the initial conditions. You're probably going to get a system of equations. What do you get?
c1 + c2 = 0
c1 = - c2
There's one equation. You've applied the x(0) = 0 condition. But you also have the other condition: x'(0) = 0. What does that give you?
whats the other condition ?
Well, you've got the function x(t). What is x'(t)?
dx/dt = c1 2.e^2t + c2 . e^t
0 = 2 c1 + c2
Excellent. So, solving those two equations simultaneously should give you c1 and c2. What do you get?
i got c1 as 0 which means c2 also 0
Thank you very much for all your help......
I dont hate the subject after all..
I going to the hostel tomorow..... I dont like it there. The food is terrible
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