Answer:
[tex](\frac{dy}{dt})^2=sin^2t[/tex]
Step-by-step explanation:
[tex]y=cost[/tex]
DE :[tex](\frac{dy}{dt})^2=1-y^2[/tex]
If y is a solution of given DE then it satisfied the DE.
Differentiate w.r.t t
[tex]\frac{dy}{dt}=-sint[/tex]
Using the formula
[tex]\frac{d(cosx)}{dx}=-sinx[/tex]
LHS:[tex](\frac{dy}{dt})^2=(-sint)^2=sin^2t[/tex]
RHS
[tex]1-y^2=1-cos^2t=sin^2t[/tex]
By using the formula
[tex]sin^2t=1-cos^2t[/tex]
LHS=RHs
Hence, y is a solution of given DE
[tex](\frac{dy}{dt})^2=sin^2t[/tex]
