unfortunately when i shud have learnt this i did'nt...after
WBJEE i learnt this...could have easily got 3 more marks..:(
What is the general solution to the differential eqn
100 d2ydx2 -20 dydx + y =0
Let y = emx be a trial solution.
so dydx = memx & d2ydx2 = m2emx
now plugging in the values of dy/dx and d2y/dx2 in the given equation we get...
100 m2emx - 20 memx + emx = 0
so emx( 100 m2 - 20m + 1) = 0
emx ≠0
so 100 m2 - 20m + 1 = 0
or ( 10m - 1)2 = 0
so m = 110 , 110
i.e. the roots of the above quadratic are real and equal. **
so general solution of the given differential equation is :
y = ( A+ Bx)(1/10)x
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** a little info....
for the general equation m2 + P1 m + P2 = 0 (called the auxillary equation)
let m1 and m2 be the roots...
CASE I
when roots are REAL AND UNEQUAL
the general solution is:
y = Aem1x + Bem2x
where A and B are two independent arbitary constants.
CASE II
when roots are REAL AND EQUAL
The general solution is :
y = (A + Bx)emx
where A and B are two independent arbitary constants.
and m = m1 = m2 (in other words m is the equal root)
CASE III
when roots are CONJUGATE COMPLEX QUANTITIES
the general solution is y = Ae(alpha)xcos{((beta)x) + B}
where A and B are two independent arbitary constants
and m1 = (alpha) + i (beta)
and m2 = (alpha) - i (beta)
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unfortunately when i shud have learnt this i did'nt...after
WBJEE i learnt this...could have easily got 3 more marks..:(