Question

The general solution of the differential equation $x + y \frac{d y}{d x} = \sec(x^2 + y^2)$ is

• A. $\sin(x^2 + y^2) = 2x + c$
• B. $\sin(x^2 + y^2) + 2x = c$
• C. $\sin(x^2 + y^2) + x = c$
• D. $\cos(x^2 + y^2) = 2x + c$

Hint:

Recognizing the derivative of the inner function inside the differential equation saves time! The term $x + y \frac{d y}{d x}$ is exactly $\frac{1}{2} \frac{d}{dx}(x^2+y^2)$. Integrating $\frac{1}{\sec u}$ (which is $\cos u$) leads straight to $\sin u$, steering you immediately to a sine-based solution!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a first-order differential equation where variables are grouped inside a trigonometric secant function as $x^2 + y^2$. We need to find its general solution.

Step 2: Key Formula or Approach:
We can use a substitution method to convert the equation into a separable form. Let's substitute the inner algebraic group: $$ \text{Put } x^2 + y^2 = u $$ Differentiating both sides with respect to $x$: $$ 2x + 2y\frac{d y}{d x} = \frac{d u}{d x} \implies x + y\frac{d y}{d x} = \frac{1}{2}\frac{d u}{d x} $$

Step 3: Detailed Explanation:
Substitute these components back into our original differential equation: $$ \frac{1}{2}\frac{d u}{d x} = \sec u $$ Separate the variables by moving all terms containing $u$ to the left and terms containing $x$ to the right: $$ \frac{d u}{\sec u} = 2 d x $$ Since $\frac{1}{\sec u} = \cos u$, the equation simplifies nicely to: $$ \cos u \ d u = 2 \ d x $$ Integrating both sides: $$ \int \cos u \ d u = \int 2 \ d x $$ $$ \sin u = 2x + c $$ Replacing $u$ with its original expression $x^2 + y^2$: $$ \sin(x^2 + y^2) = 2x + c $$

Step 4: Final Answer:
The general solution is $\sin(x^2 + y^2) = 2x + c\boxext$, matching option (A).