Differential EquationHard
Question
A function f(x) satisfying 
f (tx)dt = nf (x), where x > 0, is -
f (tx)dt = nf (x), where x > 0, is -Options
A.f (x) = c.x1-n/n
B.f (x) = c.xn/1-n
C.f (x) = c.x1/n
D.f (x) = c.x(1-n)
Solution
f(tx)dt = nf(x)Let tx = u ⇒ dt =
∴
f(u)du = nf (x) ⇒ 
f(u)du = nxf (x)f (x) = n[f (x) + xf′(x)] ⇒ f(x)
= xf′(x)
lnx = lny + lncx1-n/n = cy ⇒ y = c′x1-n/n
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