Question

Solve the following differential equation:

`("x + a")"dy"/"dx" - 3"y" = ("x + a")^5`

Answer 1

`(x + a)dy/dx - 3y = (x + a)^5`

∴ `dy/dx - 3y/(x + a) = (x + a)^4`

∴ `dy/dx + ((- 3)/(x + a))y = (x + a)^4`  ...(1)

This is the linear differential equation of the form

`dy/dx + "P".y = "Q"`, where P = `(- 3)/(x + a)` and Q = (x + a)⁴ 

∴ I.F. = `e^(int P dx) = e^(int (- 3)/(x + a)dx) = e^(-3 int 1/(x + a)dx)`

`= e^(- 3 log |x + a|) = e^(log (x + a)^- 3)`

`= (x + a)^-3 = 1/(x + a)^3`

∴ the solution of (1) is given by

`y . ("I.F.") = int "Q" * ("I.F.") dx + c`

∴ `y * 1/(x + a)^3 = int (x + a)^4 * 1/(x + a)^3 dx + c`

∴ `y/(x + a)^3 = int (x + a) dx + c`

∴ `y/(x + a)^3 = ((x + a)^2)/2 + c`

∴ 2y = (x + a)⁵ + 2c (x + a)³ 

This is the general solution.