Form the differential equations by elimniating the arbitary constants from the following equations :
(1) ` y =c^(2)+c/x` (2) ` x^(3) +y^(3) = 4ax` (3) ` y = Ae^(5x) + Be^(-5x) `
(4) ` y = A cos alpha x + B sin alpha x`

Explanation

(1) ` y = c^(2) + c/x`
differentiating w.r.t x, we get
`(dy)/(dx) = d/(dx) (c^(2) + c/x) = 0 + c( - 1/x^(2))`
` (dy)/(dx) = - c/x^(2)`
` c= x^(2) (dy)/(dx)`
Solsitiuting the value of c in (1) , we get
` y = [ -x^(2) (dy)/(dx)]^(2) + 1/x (-x^(2)(dy)/(dx))`
` x^(4) ((dy)/(dx))^(2) -x(dy)/(dx) -y =0`
This is the required D.E.
(2) The given equation is
` x^(3) +Y^(2) =4ax`
Differentiating w.r.t x, we get
` 3x^(2) '+3y^(2) (dy)/(dx) = 4a`
Subsitituting the value of 4a in (1), we get
`x^(3) +y^(2) = (3x^(2) + 3y^(2) (dy)/(dx))x`
`x^(3) +y^(3) =3x^(3) +3xy^(2) (dy)/(dx)`
`2x^(3) -y^(3) +3xy^(2) (dy)/(dx) =0`
This is the required D.E.
(3) ` y= Ae^(5x) + Be^(-5x)`
Differentiating twice w.r.t. x, we get
`(dy)/(dx) = Ae^(5x) xx 5 + Be^(-5x) xx (-5)`
`(dy)/(dx) = 5Ae^(5x) -5Be^(-5x)`
` and (d^(2)y)/(dx^(2)) =5Ae^(5x) xx 5 - 5Be^(-5x) xx (-5)`
` = 25 Ae^(5x) + 25Be^(-5x) = 25(Ae^(5x) + Be^(-5x))`
= 25 y
` (d^(2)y)/(dx^(2)) - 25y=0`
This is the required D.E.
(4) ` y = A cos alpha x + B sin alphax`
Differentiating twice w.r.t. x, we get
` (dy)/(dx)=A (- sin alphax) xx alpha + B cos alpha x xx alpha `
`(dy)/(dx) =- alpha A sin alphax + alpha B cos alpha x`
`and (d^(2)y)/(dx^(2)) =- alpha A cos alphax xx alpha + alpha B( - sin alphax)xx alpha`
`(d^(2)y)/(dx^(2)) =- alpha^(2) A cos alpha x -alpha^(2) B sin alphax`
`= - alpha^(2) (A cos alphax + B sin alphax) = - alpha^(2) y `
`(d^(2)y)/(dx^(2)) alpha^(2) y=0`
The is the reqired D.E.