Solution of the differential equation $\frac{dy}{dx} = (4x+2y+1)^2$ is ______

Question

TuteeHUB · Accepted Answer

$\frac{1}{\sqrt{2}} cot^{-1}⁡(4x+2y+1)=x+c  $

TuteeHUB · Answer

$\frac{1}{2\sqrt{2}} tan^{-1}⁡(\frac{4x+2y+1}{\sqrt{2}})=x+c  $

TuteeHUB · Answer

$\frac{1}{\sqrt{2}} cot^{-1}⁡(4x+2y+1)=x+c  $

TuteeHUB · Answer

$\frac{1}{\sqrt{2}} tan^{-1}⁡⁡(\frac{4x+2y+1}{\sqrt{2}})=c  $

TuteeHUB · Answer

cot-1⁡⁡⁡(4x+2y+1)=x+c

1.	Solution of the differential equation \(\frac{dy}{dx} = (4x+2y+1)^2\) is ______
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
Answer» B. \(\frac{1}{\sqrt{2}} cot^{-1}⁡(4x+2y+1)=x+c \)

Solution of the differential equation \(\frac{dy}{dx} = (4x+2y+1)^2\) is ______