Differential Equations!!! HELP!!!?

Question

solve following diff, equation: x^3y'=y^2 with y(1)=1

I dont know where to start!

Thanks industrie but when i rearrange dont i multiply through by the -1 thus making the c negative?
ie (y^-1)/-1= (x^-2)/-2 + c
Multiply by -1 to get y^-1 = (x^-2)/2 - c?
Is this correct?

Answer 1

Separation of variables

x^3.dy/dx=y^2

Move y to the left and x to the right
dy/y^2 = dx/x^3

Integrate (don't forget the constant of integration)
y^-2.dy=x^-3.dx
-1/y= -(1/2)x^-2 + c

1/y= (1-2cx^2)/(2x^2)
y=2x^2/(1-2cx^2)

When x=1, y=1
1=2/(1-2c)
2=1-2c
c= -1/2

y=2x^2/(1+x^2)

Answer 2

x^3y'=y^2

x^3 (dy/dx) = y^2
rearrange variables
dy/ y^2 = dx / x^3
y^-2 dy = x^-3 dx
integrate both sides

y^-1 / -1 = x^-2 / -2 + c
where c is a constant
rearrange equation
y = 2x^2 + C
sub y(1)=1
1 = 2(1)^2 + C
C = -1

then
y = 2x^2 -1

Answer 3

x³ dy/dx = y²
∫ (1/y²) dy = ∫(1/x³) dx
∫ y^(-2) dy = ∫x^(-3) dx
y^(- 1) / (- 1) = x^(-2) / (-2) + C
x^(-2) / 2 = y^(-1) + C
1 / (2x²) = 1 / y + C
y = 2x² + 2C x² y
1 = 2 + 2C
C = (- 1/2)
y = 2x² - x²y
(1 + x²).y = 2x²
y = 2x² / (1 + x²)