Solution of the differential equation \(\frac{dy}{dx} = (4x+2y+1)^2\) is ______
Posted: Thu Jul 14, 2022 9:29 am
a) \(\frac{1}{2\sqrt{2}} tan^{-1}(\frac{4x+2y+1}{\sqrt{2}})=x+c \)
b) \(\frac{1}{\sqrt{2}} cot^{-1}(4x+2y+1)=x+c \)
c) \(\frac{1}{\sqrt{2}} tan^{-1}(\frac{4x+2y+1}{\sqrt{2}})=c \)
d) cot-1(4x+2y+1)=x+c
b) \(\frac{1}{\sqrt{2}} cot^{-1}(4x+2y+1)=x+c \)
c) \(\frac{1}{\sqrt{2}} tan^{-1}(\frac{4x+2y+1}{\sqrt{2}})=c \)
d) cot-1(4x+2y+1)=x+c