Question

Solve the following Linear differential equation:

`("d"y)/("d"x) + y/((1 - x)sqrt(x)) = 1 - sqrt(x)`

Answer 1

The given linear differential equation is of the form

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

Where P = `1/((1 - x)sqrt(x))`

Q = `1 - sqrt(x)`

Take `sqrt(x)` = t

⇒ t² = x

`x^(1/2)` = t

`1/2 x^(1/2 - 1)  "d"x` = dt

`1/2 x^((-1)/2)  "d"x` = dt

`1/(2x^(1/2))  "d"x` = dt

`1/(2sqrt(x))  "d"x` = dt

`("d"x)/sqrt(x)` = 2dt

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

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

= `"e"^(int 1/(1 - "t"^2) 2 "dt")`

= `"e"^(int (2"dt")/(1 - "t"^2)`

= `"e"^(log((1 + "t")/(1 - "t"))` 

` ∵ int ("d"x)/("a"^2 - x^2) = 1/(2x) log |("a" + x)/("a" - x)`

a = 1, x = t

`1/(2x) log|("a" + x)/("a" - x)| = 2 xx 1/2 log((1 + "t")/(1 - "t"))`

I.F = `(1 + "t")/(1 - "t")`

I.F = `(1 + sqrt(x))/(1 - sqrt(x))`

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

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

= `int (1 + sqrt(x)) "d"x + "c"`

`y((1 + sqrt(x))/(1 - sqrt(x))) = x + x^(3/2)/(3/2) + "c"`

`y((1 + sqrt(x))/(1 - sqrt(x))) = x + 2/3 x^(3/2) + "c"`

`y((1 + sqrt(x))/(1 - sqrt(x))) = x + 2/3 x sqrt(x) + "c"`

∵ `x^(3/2) = xsqrt(x)`