Question

Solve the following Linear differential equation:

`(x^2 + 1) ("d"y)/("d"x) + 2xy = sqrt(x^2 + 4)`

Answer 1

The given differential equation may be written as

`((x^2 + 1)/(x^2 + 1)) ("d"y)/("d"y) + ((2x)/(x^2 + 1))y = sqrt(x^2 + 4)/(x^2 + 1)`

⇒ `("d"y)/("d"y) + ((2x)/(x^2 + 1))y = sqrt(x^2 + 4)/(x^2 + 1)`

This is of the form `("d"y)/("d"x) + "P"y` = Q

Where P = `(2x)/(x^2 + 1)`

Q = `sqrt(x^2 + 4)/(x^2 + 1)`

Thus, the given differential equation is linear.

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

= `"e"^(int (2x)/(x + 1)"d"x)`

= `"e"^(log(x^2 + 2))`

= x² + 1

So, the required solution is given by

`y xx "I.F" = int ("Q" xx "I.F")  "d"x + "c"`

`y(x^2 + 1) = int sqrt(x^2 + 4)/(x^2 + 1) xx (x^2 + 1)  "d"x`

`y(x^2 + 1) = int sqrt(x^2 + 4)  "d"x`  ........`[because int sqrt(x^2 + "a"^2)  "d"x = 1/2 x sqrt(x^2 + "a"^2) + "a"^2/2 log |x + sqrt(x^2 - "a"^2)| + "c"]`

`y(x^2 + 1) = 1/2 x sqrt(x^2 + 4) + 1/2 xx 2^2 xx log|x + sqrt(x^2 + 4)| + "c"`

`y(x^2 + 1) = x/2 sqrt(x^2 + 4) + 2 log|x + sqrt(x^2 + 4)| + "c"`

Hence `y(x^2 + 1) = x/2 sqrt(x^2 + 4) + 2 log|x + sqrt(x^2 + 4)| + "c"` is the required solution.