First order differential equations (integrating factor)
4
By using an integrating factor, find the general solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 4 } { 2 x + 1 } y = 4 ( 2 x + 1 ) ^ { 5 }$$
giving your answer in the form \(y = \mathrm { f } ( x )\).
The gradient of a curve at any point \(( x , y )\) on the curve is given by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 ( 2 x + 1 ) ^ { 5 } - \frac { 4 } { 2 x + 1 } y$$
The point whose \(x\)-coordinate is zero is a stationary point of the curve. Using your answer to part (a), find the equation of the curve.