Question

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

• A. \( y = \frac{1}{2} \tan^{-1}(x + y) + c \)
• B. \( y = -(x + y)^{-1} + c \)
• C. \( y = \frac{1}{3}(x + y)^3 + c \)
• D. \( y = \sin^{-1}(x + y) + c \)
• E. \( y = \tan^{-1}(x + y) + c \)

Hint:

When dealing with a separable differential equation, always remember to separate the variables first before integrating. If needed, use substitution to simplify integrals.

Answer 1

Answer: (E)

Step 1: Start with the given differential equation: \[ (x + y)^2 \frac{dy}{dx} = 1. \] Rearrange to separate variables: \[ \frac{dy}{(x + y)^2} = \frac{dx}{1}. \] Step 2: Integrate both sides: \[ \int \frac{dy}{(x + y)^2} = \int dx. \] Step 3: The integral on the left-hand side can be solved by substituting \( u = x + y \), so \( du = dx \). This gives: \[ \int \frac{du}{u^2} = \int dx. \] The integral of \( \frac{1}{u^2} \) is \( -\frac{1}{u} \), so: \[ -\frac{1}{x + y} = x + c. \] Step 4: Simplify the equation: \[ \frac{1}{x + y} = -(x + c). \] Step 5: Now, take the inverse of both sides: \[ x + y = \frac{1}{-(x + c)}. \] Therefore, the solution to the differential equation is: \[ y = \tan^{-1}(x + y) + c. \]