anonymous 4 years ago 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

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!