अवकल समीकरण (x+y)dx + [3y² + 2y + (x+y)]dy = 0 का हल होगा : | Agrigyan

The given differential equation is of the form $M\,dx + N\,dy = 0$ with:
$M = \cos(x+y)$,
$N = 3y^2 + 2y + \cos(x+y)$.
Check for exactness:
$\frac{\partial M}{\partial y} = -\sin(x+y)$,
$\frac{\partial N}{\partial x} = -\sin(x+y)$.
Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.
The solution is obtained by:
$\int M\,dx$ (treating $y$ as constant) $+ \int (\text{terms in } N \text{ free from } x)\,dy = C$.
$\int \cos(x+y)\,dx = \sin(x+y)$.
Terms of $N$ free from $x$: $3y^2 + 2y$, so $\int (3y^2 + 2y)\,dy = y^3 + y^2$.
Combining gives:
$\sin(x+y) + y^3 + y^2 = C$.

Detailed analysis of options:

• $\sin(x+y) + y^3 + y^2 = C$ is correct.
• Has $x-y$ instead of $x+y$.
• Replaces $y^3 + y^2$ with $x^3 + x^2$.
• Has $x-y$ and $x$ terms.