Question

Solve the following Linear differential equation:

`(x + "a") ("d"y)/("d"x) - 2y = (x + "a")^4`

Answer 1

`(x + "a") ("d"y)/("d"x) - 2y = (x + "a")^4`

`((x + "a"))/((x + "a")) ("d"y)/("d"x) - (2y)/(x + "a") = (x + "a")^4/((x + "a"))`

÷ By (x + a) on both sides,

`("d"y)/("d"x) - (2"y")/(x + "a") = (x + "a")^3`

This is of the form

`("d"y)/("d"x) + "P"y` = Q

Where P = `(-2)/(x + "a")`

Q = `(x + "a")^3`

`int "Pd"x = int (-2)/(x + "a")  "d"x`

= `- 2 int ("d"x)/(x + "a")`

= `- 2 log(x + "a")`

= log(x + "a")^-2`

`int "Pd"x = log  1/(x + "a")^2`

I.F = `"e"^(int "Pd"x)`

= `"e"^(log  1/(x + "a")^2)`

= `1/(x + "a")^2`

The solution is `y xx "I.F" = int "Q" xx "I.F"  "d"x + "c"`

`y/(x + "a")^2 = int (x + "a")  "d"x + "c"`

`y/(x + "a")^2 = (x + "a")^2/2 + "c"`

`y/(x + "a")^2 = ((x + "a")^2 + 2"c")/2`

2y = `(x  + "a")^2 [(x + "a")^2 + 2"c"]`

2y = `(x + "a")^2 + 2"c"  (x + "a")^2` is a required solution.