Question

Solve \( x + y \frac{dy}{dx} = \sec (x^2 + y^2) \)

Hint:

For equations involving \( x^2 + y^2 \), try substitution \( u = x^2 + y^2 \) to simplify.

Answer 1

Let \( u = x^2 + y^2 \). 
\[ \frac{du}{dx} = 2x + 2y \frac{dy}{dx} \Rightarrow y \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx} - x. \] Substitute: 
\[ x + \left( \frac{1}{2} \frac{du}{dx} - x \right) = \sec u \Rightarrow \frac{1}{2} \frac{du}{dx} = \sec u \Rightarrow \frac{du}{dx} = 2 \sec u. \] \[ \int \cos u \, du = \int 2 \, dx \Rightarrow \sin u = 2x + c. \] \[ \sin (x^2 + y^2) = 2x + c. \] Answer: \( \sin (x^2 + y^2) = 2x + c \).