Question

solve the given differential equation.
(9x + 1)y^{2}\frac{dy}{dx}+4x^{2}+3y^{3}=0

Explanation:

Step1: Rearrange the differential equation

Separate the variables: $(9x + 1)y^{2}\frac{dy}{dx}=-4x^{2}-3y^{3}$. Then, $\frac{y^{2}}{-4x^{2}-3y^{3}}dy=\frac{1}{9x + 1}dx$.

Step2: Integrate both sides

Integrate $\int\frac{y^{2}}{-4x^{2}-3y^{3}}dy=\int\frac{1}{9x + 1}dx$. Let $u=-4x^{2}-3y^{3}$, then $du=-9y^{2}dy$ for the left - hand side integral and let $v = 9x+1$, $dv=9dx$ for the right - hand side integral.
The left - hand side integral becomes $-\frac{1}{9}\int\frac{du}{u}$ and the right - hand side becomes $\frac{1}{9}\int\frac{dv}{v}$.

Step3: Evaluate the integrals

$-\frac{1}{9}\ln|u|=\frac{1}{9}\ln|v|+C$. Substituting back $u=-4x^{2}-3y^{3}$ and $v = 9x + 1$, we get $-\ln|-4x^{2}-3y^{3}|=\ln|9x + 1|+C$.
Using the property of logarithms $a\ln b=\ln(b^{a})$ and $\ln a+\ln b=\ln(ab)$, we can rewrite it as $\ln|(9x + 1)(-4x^{2}-3y^{3})|=-C$. Let $K = e^{-C}$, then $(9x + 1)(-4x^{2}-3y^{3})=K$.

Answer:

$(9x + 1)(-4x^{2}-3y^{3})=K$ (where $K$ is an arbitrary constant)