Question

Solve the following Linear differential equation:

`(1 - x^2) ("d"y)/("d"x) - xy` = 1

Answer 1

The given differential equation can be written as

`(1 - x^2)/(1 - x^2) ("d"y)/("d"x) - x/(1 - x^2) y = 1/(1 - x^2)`

`("d"y)/("d"x) - (x/(1 - x^2))y = 1/(1 - x^2)`

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

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

Q = `1/(1 - x^2)`

Thus, the given differential equation is linear.

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

= `"e"^(int (-x)/(1 - x^2)  "d"x)`

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

= `"e"^(1/2 log(1 - x^2))`

= `"e"^(logsqrt(1 - x^2))`

= `sqrt(1 - x^2)`

So, the required solution is given by

`[y xx "I.F"] = int["Q" xx "I.F"] "d"x + "c"`

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

= `int 1/sqrt(1 - x^2) sqrt(1 - x^2)/sqrt(1 - x^2)  "d"x + "c"`

= `int 1/sqrt(1 - x^2)  "d"x + "c"`

`ysqrt(1 - x^2) = sin^-1x + "c"`

`÷ sqrt(1 - x^2), (ysqrt(1 - x^2))/(sqrt(1 - x^2)) = (sin^-1x)/(sqrt(1 - x^2)) + "c"/sqrt(1 - x^2)`

y = `(sin^-1x)/sqrt(1 - x^2) + "c"(1 - x^2)^((-1)/2)`

Which is a required solution.