The solution of differential equation ` t = 1 + (ty)(dy)/(dt) + (ty)^2/(2!) ((dy)/(dt))^(2)+ . . . . oo ` is

Options

A. ` y = pm sqrt((logt)^(2)+C)`
B. ` ty = t^(y) +C`
C. ` y = log t +C`
D. ` y = (log t)^(2)+C`

Answer

a

Explanation

The given equation is ` t = 1 + (ty)((dy)/(dt)) + ((ty)^(2))/(2!) ((dy)/(dt))^(2) + …oo`
` rArr t = e^(ty((dy)/(dt))) " " [ :' e^(x) = 1+ x + (x^(2))/(2!)+...]`
`rArr log t = ty (dy)/(dt) rArr y dy = (log t)/t dt`
On integrating both sides , we get
` (y^(2))/2 = ((logt)^(2))/2 = k `
` rArr y = pm sqrt((logt)^(2) +2k) rArr y = pm sqrt((logt)^(2)+C)`