Question

Find the specific solution of the differential equation dy/dx = 4y/x^2 with condition y(−4) = e.

y= -1 - 4/x <- my answer
y=-1*e^(1/x)
y=e^(-4/x)
None of these

Answer 1

Nope. That condition y(-4) = e should have tipped you off that logs/exponentials were involved.

dy/dx = 4y/x^2
dy/y = 4/x^2
lny = -4/x + c
y = c*e^(-4x)

note that the two c values are different. We could have said that

lny = -4/x + lnc
y = e^(-4/x + lnc)
= e^(-4/x)*e^(lnc)
= c*e^(-4/x)

Now, y(-4) = e, so

ce^(-4/-4) = e
ce^1 = e
c = 1

y = e^(-4/x)

Did you check your answer?

y = -1 - 4/x
y' = 4/x^2
But you wanted y' = 4y/x^2

checking my answer,

y' = (4/x^2) e^(-4/x) = 4y/x^2