Solving differential equations with negative exponents

[ucci]

Find an equation of the curve that passes through the point (1,1)
(dy/dx)= (y^2/x^3)

This what I've done so far and also where I get stuck. please help.
DX (1/ X^3) = DY(1/Y^2)
DX X^-3= DY Y^-2
(-1/2) X^-2= (-Y)^ -1

 

[galactus]

You could separate variables and get $\displaystyle \frac{dy}{y^{2}}=\frac{dx}{x^{3}}$

Integrate: $\displaystyle \int\frac{1}{y^{2}}dy=\int\frac{1}{x^{3}}dx$

$\displaystyle \frac{-1}{y}=\frac{-1}{2x^{2}}+C$

$\displaystyle y=\frac{-2x^{2}}{2Cx^{2}-1}$

Now, solve for C by using the (1,1) and you have it kicked.

 

[Deleted member 4993]

Find an equation of the curve that passes through the point (1,1)
(dy/dx)= (y^2/x^3)

This what I've done so far and also where I get stuck. please help.
DX (1/ X^3) = DY(1/Y^2)
DX X^-3= DY Y^-2
(-1/2) X^-2= (-Y)^ -1

You have done everything okay - except you forgot your constant of integration