Question

The solution of the differential equation $\cos(x+y)dx + [3y^2 + 2y + \cos(x+y)]dy = 0$ is given by :

• A. $\sin(x+y) + y^3 + y^2 = \text{C}$
• B. $\sin(x-y) + y^3 + y^2 = \text{C}$
• C. $\sin(x+y) + x^3 + x^2 = \text{C}$
• D. $\sin(x-y) + x^3 + x^2 = \text{C}$

Hint:

If an equation is exact, the solution is always the sum of the integral of $M$ (with respect to $x$) and the integral of any purely $y$-dependent terms in $N$. This shortcut bypasses lengthy integrations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a first-order differential equation of the form: \[ M(x, y) \, dx + N(x, y) \, dy = 0 \] We first test if the equation is exact by verifying if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
Key Formula or Approach:
If the equation is exact, the general solution is: \[ \int M \, dx \quad (\text{keeping } y \text{ constant}) + \int (\text{terms in } N \text{ free from } x) \, dy = C \]

Step 2: Detailed Explanation:
From the given differential equation: \[ M = \cos(x + y) \] \[ N = 3y^2 + 2y + \cos(x + y) \] Let us find the partial derivatives: \[ \frac{\partial M}{\partial y} = -\sin(x + y) \] \[ \frac{\partial N}{\partial x} = -\sin(x + y) \] Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the differential equation is exact.
Now, integrate $M$ with respect to $x$, keeping $y$ as a constant: \[ \int M \, dx = \int \cos(x + y) \, dx = \sin(x + y) \] Next, identify the terms in $N$ that do not contain the variable $x$: \[ \text{Terms free from } x = 3y^2 + 2y \] Integrate these terms with respect to $y$: \[ \int (3y^2 + 2y) \, dy = y^3 + y^2 \] Combine the two results to write the general solution: \[ \sin(x+y) + y^3 + y^2 = \text{C} \]

Step 3: Final Answer:
The solution is $\sin(x+y) + y^3 + y^2 = \text{C}$, matching Option (A).