solution:
Consider the differential equation,
7ty + (1+t2)1/2y1 = 0
Rewrite the DE as,
(1+t2)1/2dy = - 7ty
dy/y = -7t/√1+t2 dt
in y = -7(1+t2) + c
y = ce-7(1+t2)
given,
y(0) = 1 => ce-7 = -1 => c = e7
ᴪ (t,y) = y -ce ᴪ(0,1)(1+t2)
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