Question

solve the initial value problem, xy'=y+x^3, y(2)=5.

Answer 1

xy'=y+x^3
xdy/dx -y =x^3
xdy-ydx =x^3 dx
(xdy-ydx)/X^2 =xdx
d(y/x) = xdx...
now integrate both sides...

Answer 2

integrating we have
y/x = x^2/2 +c

Answer 3

what about y(2)=5？

Answer 4

and how does xdy-ydx become d(y/x) instead of d(x/y)? and why does x^2 disappear?

Answer 5

you need to find c

Answer 6

put x=2 and y=5 to find c