Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
Is this correct?
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)'=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)=\int \frac{3x^2}{{(2x+1)}^{\frac{3}{2}}} \mathop{dx}$$

$$\frac{dy}{dx}  -\frac{3}{2x + 1}y = 3x^2$$
For an integrating factor we have
$$\frac{dy}{dx} + Py = Q, \quad I = \exp(\int P \; dx)$$
then $$Iy = \int IQ\;dx$$
For our method
$$I = \exp(\int -\frac{3}{2x + 1} \; dx) = \exp(-\frac{3}{2}ln(2x + 1)) = (2x + 1)^{-\frac{3}{2}}$$
so
$$y = (2x + 1)^\frac{3}{2} \int \frac{3x^2}{(2x + 1)^{\frac{3}{2}}} \; dx= C_1(2x + 1)^\frac{3}{2} + (2x + 1)(x^2 - 2x -2)$$
since the integral gives
$$\frac{x^2 - 2x - 2}{(2x + 1)^\frac{1}{2}}$$
