Divide by $(2x^3 + C)$: $$ y = \frac{-1}{2x^3 + C} $$
∫y-2dy=y-1-1=−1yintegral of y to the negative 2 power space d y equals the fraction with numerator y to the negative 1 power and denominator negative 1 end-fraction equals negative 1 over y end-fraction Use the same rule for x2x squared solve the differential equation. dy dx = 6x2y2
where (C) is an arbitrary constant.
We multiply both sides by $dx$ and divide both sides by $y^2$. Divide by $(2x^3 + C)$: $$ y =
Using the power rule for $x$: $$ \int 6x^2 , dx = 6 \left( \frac{x^3}{3} \right) = 2x^3 $$ The left side integrates to $-\frac{1}{y}$ and the
Now, integrate both sides of the equation: $\int \frac{dy}{y^2} = \int 6x^2 dx$. The left side integrates to $-\frac{1}{y}$ and the right side to $2x^3 + C$, where $C$ is the constant of integration.