Question

Which of the following equations is exact?

• A. \(x \, dy + y^2 \, dx = 0\)
• B. \((1/4)e^{4y} \, dx + x e^{4y} \, dy = 0\)
• C. \((\cos y + x^2) \, dx - x \csc^2 y \, dy = 0\)
• D. \(x \, dy - y \, dx - \cos(1/x) \, dx = 0\)

Hint:

Exam Tip:
To check if a differential equation is exact:

• Write it as \(M \, dx + N \, dy = 0\).
• Check if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A differential equation \(M(x, y) \, dx + N(x, y) \, dy = 0\) is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

Step 2: Checking Each Option:

• Option (A): \(M = y^2, N = x\).
\(\frac{\partial M}{\partial y} = 2y\), \(\frac{\partial N}{\partial x} = 1\).
Not equal. So, not exact.
• Option (B): \(M = \frac{1}{4}e^{4y}, N = x e^{4y}\).
\(\frac{\partial M}{\partial y} = \frac{1}{4} \cdot 4 e^{4y} = e^{4y}\).
\(\frac{\partial N}{\partial x} = e^{4y}\).
Equal! So, this is exact.
Wait, let's check: \(\frac{\partial N}{\partial x} = e^{4y}\).
Yes, they are equal. So, option (B) is exact.
• Option (C): \(M = \cos y + x^2, N = -x \csc^2 y\).
\(\frac{\partial M}{\partial y} = -\sin y\).
\(\frac{\partial N}{\partial x} = -\csc^2 y\).
Not equal (unless \(\sin y = \csc^2 y\), which is not generally true). So, not exact.
• Option (D): \(M = -y - \cos(1/x), N = x\).
\(\frac{\partial M}{\partial y} = -1\).
\(\frac{\partial N}{\partial x} = 1\).
Not equal. So, not exact. So, option (B) is exact. But the answer key says option (C).
Let's re-evaluate option (C) carefully: \[ (\cos y + x^2) \, dx - x \csc^2 y \, dy = 0 \] \(M = \cos y + x^2\), \(N = -x \csc^2 y\).
\(\frac{\partial M}{\partial y} = -\sin y\).
\(\frac{\partial N}{\partial x} = -\csc^2 y\).
These are not equal. So, (C) is not exact.
Option (B) is exact. Let's verify: \[ \frac{1}{4} e^{4y} \, dx + x e^{4y} \, dy = 0 \] \(M = \frac{1}{4} e^{4y}\), \(N = x e^{4y}\).
\(\frac{\partial M}{\partial y} = \frac{1}{4} \cdot 4 e^{4y} = e^{4y}\).
\(\frac{\partial N}{\partial x} = e^{4y}\).
They are equal. So, option (B) is exact.
There is a discrepancy. The answer key says (C).
Let's check option (C) again: \(M = \cos y + x^2\), \(N = -x \csc^2 y\).
If we rewrite \(N\) as \(-x \csc^2 y\), then \(\frac{\partial N}{\partial x} = -\csc^2 y\).
But \(\frac{\partial M}{\partial y} = -\sin y\).
For exactness, we need \(-\sin y = -\csc^2 y \Rightarrow \sin y = \csc^2 y \Rightarrow \sin^3 y = 1\), which is not true for all y.
So, (C) is not exact.
I'll go with option (B) as the correct answer.

Step 4: Final Answer:
Therefore, option (B) is correct.