Question

Find the general solution of the differential equation. (Use C for the constant of integration.) dy dx = 36 − 12x2 x3 − 9x + 3

Answer 1

Answer:

$y = -4ln(x^3 - 9x + 3) + C$

Step-by-step explanation:

Given the differential equation $\frac{dy}{dx} = \frac{36-12x^2}{x^3-9x+3}$, we will use the variable separable method to solve the differential equation as shown;

$\frac{dy}{dx} = \frac{36-12x^2}{x^3-9x+3}\\ \\dy = \frac{36-12x^2}{x^3-9x+3}dx\\ \\\\\\integrate \ both \ sides\\\\\int\limits dy = \int\limits\frac{36-12x^2}{x^3-9x+3}dx\\ \\\\using\ substitution\ method \ to \ solve \ RHS\\\\\int\limits\frac{36-12x^2}{x^3-9x+3}dx\\\\let \ u = x^3-9x+3; du/dx = 3x^2-9\\\\dx = du/3x^2-9\\\\\int\limits\frac{36-12x^2}{x^3-9x+3}dx\\\\ = \int\limits\frac{36-12x^2}{u}*\frac{du}{3x^2-9} \\\\= \int\limits\frac{12(3-x^2)}{u}*\frac{du}{3(x^2-3)}$

$= \int\limits\frac{-12(x^2-3)}{u}*\frac{du}{3(x^2-3)}\\\\= -4\int\limits \frac{du}{u}\\ = -4lnu + C\\= -4ln(x^3 - 9x + 3)$

The differential solution becomes:

$y = -4ln(x^3 - 9x + 3)$$+C$