What is the only solution of the initial value problem `y^(2)=t(1+y) ,y(0)=0` ?

Options

A. `y=-1+e^(1^(2)//2`
B. `y=1+e^(t^(2)//2)`
C. y=-t
D. y=t

Answer

A

Explanation

Given equation is : y' = t ( 1+y)
i.e, ` (dy)/(dt)= t (1+y)`
`Rightarrow int 1/(1+y) dy = int tdt Rightarrow log ( 1 +y) = t^(2)/2 +c`
As per initial conditions
` y (0) = 0 when t =0, y =0`
` Rightarrow log 1 =c Rightarrow c =0`
`log (1 +y) = t^(2)/2 Rightarrow 1+y = e^(t^(2)//2)`
which is required solution .