Express 4x+5(x+2)(2x+1)\displaystyle \frac{4x + 5}{(x + 2)(2x + 1)}(x+2)(2x+1)4x+5 in partial fractions
Given that x>0x > 0x>0, find the general solution to the differential equation
(x+2)(2x+1)dydx=y(4x+5) (x + 2)(2x + 1) \frac{dy}{dx} = y(4x + 5) (x+2)(2x+1)dxdy=y(4x+5)Hence find the particular solution to the differential equation that satisfies y=24y = 24y=24 at x=1x = 1x=1, giving your answer in the form y=f(x)y = f(x)y=f(x)
