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Nischitha. J posted an Question
July 01, 2020 • 15:52 pm
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30 points
30 points
- IIT JAM
- Mathematics (MA)
Dy py dx (b) since u and v are the solutions of + py = q, .1) du dx (2) therefore +pu q dv .3) +pv = q dx and d ...4) subtracting (3) from (2), we get (u v) + p
dy Py dx (b) Since u and v are the solutions of + Py = Q, .1) du dx (2) therefore +Pu Q dv .3) +Pv = Q dx and d ...4) Subtracting (3) from (2), we get (u v) + P(u v) = 0. Subtracting (3) from (1), we get(y- v) + P(y- v) = 0 From (4) and (5), we have Integrating, we get log (y- v) = log(u v) + log k, where k is a constant. dx ..(5) d(y-v) y-v du-V)- P dx y-v u-V Thus y - v = k(u - v), giving y = k (u - v) + V. ndunt of the slone and the prdinate at any point (x, y) of a curve passing tn .I didn't understand the problem after 5th equation
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Nischitha. j
Thank u sir