Question

Which one of the following is a non-linear, second-order, and first-degree differential equation?

• A. \( \dfrac{d^2y}{dx^2} + \cos x \dfrac{dy}{dx} + e^x y = 0 \)
• B. \( \dfrac{d^2y}{dx^2} + \cos y = 0 \)
• C. \( \dfrac{d^2y}{dx^2} + \sin x = 0 \)
• D. \( \left(\dfrac{dy}{dx}\right)^2 + 3y = 0 \)

Hint:

Order comes from the highest derivative present, degree from its power, and linearity fails only when y itself sits inside a non-linear function.

Answer 1

The Correct Option is B

Step 1: Recall what order, degree, and linearity mean.
The order of a differential equation is the order of its highest derivative.
The degree is the power of the highest order derivative, once the equation is written as a polynomial in derivatives.
An equation is linear if the dependent variable y and all its derivatives appear only to the first power, and no term wraps y or a derivative of y inside a non-linear function of y. Coefficients can be any function of the independent variable x.

Step 2: Check option A.
\[ \frac{d^2y}{dx^2} + \cos x \frac{dy}{dx} + e^x y = 0 \]
The highest derivative is \( \frac{d^2y}{dx^2} \), so order is 2 and degree is 1.
The coefficients \( \cos x \) and \( e^x \) are functions of x only, and y and its derivatives appear only to the first power, so this equation is linear. Not the answer.

Step 3: Check option B.
\[ \frac{d^2y}{dx^2} + \cos y = 0 \]
The highest derivative is again \( \frac{d^2y}{dx^2} \), so order 2, degree 1.
But the term \( \cos y \) is a non-linear function of the dependent variable y itself, not of x, so this makes the equation non-linear.
This equation is non-linear, second order, and first degree, which is what the question asks for.

Step 4: Check option C.
\[ \frac{d^2y}{dx^2} + \sin x = 0 \]
Here \( \sin x \) is a function of the independent variable x, not of y, so it is fine in a linear equation. This equation is linear, second order, first degree, not the answer.

Step 5: Check option D.
\[ \left(\frac{dy}{dx}\right)^2 + 3y = 0 \]
The highest derivative here is \( \frac{dy}{dx} \), so order is 1, not 2.
Its power is 2, so the degree is 2. This equation is first order, second degree, and non-linear, which does not match the required second-order, first-degree condition.

Final Answer:
Only option B is non-linear, second order, and first degree.
\[ \boxed{\text{Option (B)}} \]