Answer:
[tex]y = -4ln(x^3 - 9x + 3) + C\\[/tex]
Step-by-step explanation:
Given the differential equation [tex]\frac{dy}{dx} = \frac{36-12x^2}{x^3-9x+3}\\ \\[/tex], we will use the variable separable method to solve the differential equation as shown;
[tex]\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)}\\\\[/tex]
[tex]= \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)[/tex]
The differential solution becomes:
[tex]y = -4ln(x^3 - 9x + 3)[/tex][tex]+C[/tex]
