A continuously differentiable function $y=f(x),x∈(\frac{-\pi}{2},\frac{\pi}{2})$ satisfying $y'=1+y^2,y(0)=0$ is:
Answer & explanation
Correct answer: option 1
$\int\frac{dy}{1+y^2}=\int dx⇒tan^{-1}y=x+c$;
y(0) = 0 ⇒ c = 0 ⇒ y = tan x
A continuously differentiable function $y=f(x),x∈(\frac{-\pi}{2},\frac{\pi}{2})$ satisfying $y'=1+y^2,y(0)=0$ is:
Correct answer: option 1
$\int\frac{dy}{1+y^2}=\int dx⇒tan^{-1}y=x+c$;
y(0) = 0 ⇒ c = 0 ⇒ y = tan x