1. Let a and b be coefficients such that
[tex]\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}[/tex]
Combining the fractions on the right gives
[tex]\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}[/tex]
[tex]\implies 1 = (2a+b)x + 3a[/tex]
[tex]\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23[/tex]
so that
[tex]\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}[/tex]
2. a. The given ODE is separable as
[tex]x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}[/tex]
Using the result of part (1), integrating both sides gives
[tex]\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C[/tex]
Given that y = 1 when x = 1, we find
[tex]\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)[/tex]
so the particular solution to the ODE is
[tex]\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)[/tex]
We can solve this explicitly for y :
[tex]\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)[/tex]
[tex]\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|[/tex]
[tex]\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|[/tex]
[tex]\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}[/tex]
2. b. When x = 9, we get
[tex]y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}[/tex]
