Dy/dx = y/x^2 dy/y = dx/x^2 ln y * ln c = -1/x cy = e^(-1/x) y = ce^(-1/x)

Dy/dx = y/(x^2)
dy/y = dx/(x^2)
int[dy/y] = int[dx/(x^2)] ... apply integral to both sides
ln(|y|) = (-1/x) + C
|y| = e^{(-1/x) + C}
|y| = e^C*e^(-1/x)
|y| = C*e^(-1/x)
y = C*e^(-1/x)

So you have the correct answer. Nice job.



Check:
y = C*e^(-1/x)
dy/dx = d/dx[C*e^(-1/x)]
dy/dx = d/dx[-1/x]*C*e^(-1/x)
dy/dx = (1/(x^2))*C*e^(-1/x)
is the expression for the left hand side (LHS)

y/(x^2) = [C*e^(-1/x)]/(x^2)
y/(x^2) = (1/(x^2))*C*e^(-1/x)
is the expression for the right hand side (RHS)

Since LHS = RHS, this confirms the solution for dy/dx = y/(x^2)

Bis your answer. it only holds the decimal numerals