Question

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

• A. \(y = \tan x + c\)
• B. \(y = \tan(x+c)\)
• C. \(y = \tan x\)
• D. \(y = -\tan(x+c)\)

Hint:

When solving differential equations, remember that the constant of integration \(c\) is added immediately after integrating. In this case, `tan⁻¹(y) = x + c` becomes `y = tan(x+c)`, which is different from `y = tan(x) + c`. The position of the constant is crucial.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the general solution of the given first-order differential equation.

Step 2: Key Formula or Approach:
The equation is of the variable separable type. We will separate the terms involving \(y\) and \(dy\) from the terms involving \(x\) and \(dx\), and then integrate both sides. The key integral required is \(\int \frac{1}{1+y^2} dy = \tan^{-1}(y)\).

Step 3: Detailed Explanation:
The differential equation is:
\[ \frac{dy}{dx} = 1 + y^2 \]
Separate the variables by multiplying by \(dx\) and dividing by \((1+y^2)\):
\[ \frac{dy}{1+y^2} = dx \]
Now, integrate both sides of the equation:
\[ \int \frac{1}{1+y^2} \,dy = \int 1 \,dx \]
Performing the integration gives:
\[ \tan^{-1}(y) = x + c \]
where \(c\) is the constant of integration.
To find the explicit solution for \(y\), we take the tangent of both sides:
\[ y = \tan(x+c) \]

Step 4: Final Answer:
The solution of the differential equation is \(y = \tan(x+c)\).