Question:

The solution of the differential equation \[ \frac{dy}{dx}=1+y^2 \] is

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Remember \(\int\frac{dy}{1+y^2}=\tan^{-1}y\).
  • \(y=\tan x+c\)
  • \(y=\tan(x+c)\)
  • \(y=\tan x\)
  • \(y=-\tan(x+c)\)
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The Correct Option is B

Solution and Explanation


Step 1:
Given: \[ \frac{dy}{dx}=1+y^2 \]

Step 2:
Separate variables: \[ \frac{dy}{1+y^2}=dx \]

Step 3:
Integrate both sides: \[ \int\frac{dy}{1+y^2}=\int dx \] \[ \tan^{-1}y=x+c \]

Step 4:
Taking tangent both sides: \[ y=\tan(x+c) \] \[ \boxed{y=\tan(x+c)} \]
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