Question

Solve the differential equation dy/dx= y^2/x^3 for y=f(x) with the condition y(1) = 1.

Answer 1

we will use seperation of variables

tmes both sies by 1/(y^2)
times both sides by dx

we get

1/(y^2)dy=1/(x^3)dx
integrate both sides
-1/y=1/(2x^2)+c

initial condition, (1,1)
-1/1=1/(2(1^2))+c
-1=1/2+c
-3/2=c


-1/y=1/(2x^2)-3/2
solve for y
-1=y(1/(2x^2)-3/2)
y=-1/(1/2x^2-3/2)
$y= \frac{2x^2}{3x^2-1}$