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Question

Find the specific solution of the differential equation  with condition y(−4) = 1.

Answer 1

dy/dx = 4 y / x^2

x^2 dy = 4 y dx

dy/y = 4 dx / x^2

ln y = 4 (-x ^(-1)) +c

y = exp ( -4/x + c)

initiial condition: 1 = exp ( -4/-4 + c)

1 = exp (c + 1)

ln 1 = c+1

c = -1

y = exp ( -4/x - 1)

check:

ln y = -4/x - 1

dy/y = 4/x^2 dx

which is the same as the original D.E