How would I start working a first order differential equation (with an initial value) like this?

(x^3 - y) dx + x dy = 0, y(1) = 3

Question

How would I start working a first order differential equation (with an initial value) like this?

(x^3 - y) dx + x dy = 0, y(1) = 3

anonymous · Answer

The given equation is :
$$x ^{3} + x*dy(x)/dx - y(x) = 0$$
= $$dy(x)/dx - y(x)/x  = -x ^{2}$$
Now put, $$\mu(x) = e ^{\int\limits -1/x dx} = 1/x$$
Multiply both sides of the equation by $$\mu(x)$$to get :
$$[dy(x)/dx]/x  -  y(x)/x ^{2} = -x$$
= $$[dy(x)/dx]/x +d/dx(1/x)y(x) = -x$$
Now apply the reverse product rule to get:
$$d/dx(y(x)/x) = -x$$
or,$$\int\limits d/dx(y(x)/x) dx = \int\limits -x dx$$
or,$$y(x)/x = -x ^{2}/2 +c _{1}$$
Now you can proceed easily!