Question

The solution of $(e^y + 1)\cos x\,dx + e^y \sin x\,dy = 0$ is

• A. $(e^y + 1)\sin x = C$
• B. $(e^y + 1) = C\sin x$
• C. $e^y = C\sin x$
• D. $(e^y - 1)\sin x = C$
• E. $(e^y + 1)\cos x = C$

Hint:

For exact differential equations: - Always check $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ - Integrate $M$ w.r.t $x$, then adjust using $N$

Answer 1

The Correct Option is A

Concept: This is a differential equation of the form: \[ M(x,y)\,dx + N(x,y)\,dy = 0 \] We check if it is exact: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]

Step 1: Identify $M$ and $N$.
\[ M = (e^y + 1)\cos x, \quad N = e^y \sin x \]

Step 2: Check exactness.
\[ \frac{\partial M}{\partial y} = e^y \cos x \] \[ \frac{\partial N}{\partial x} = e^y \cos x \] Since both are equal, the equation is exact.

Step 3: Find the potential function $\psi(x,y)$.
Integrate $M$ with respect to $x$: \[ \psi = \int (e^y + 1)\cos x \, dx = (e^y + 1)\sin x + h(y) \]

Step 4: Differentiate $\psi$ w.r.t $y$ and compare with $N$.
\[ \frac{\partial \psi}{\partial y} = e^y \sin x + h'(y) \] Equate with $N = e^y \sin x$: \[ h'(y) = 0 \Rightarrow h(y) = \text{constant} \]

Step 5: Write the final solution.
\[ (e^y + 1)\sin x = C \]