Question

Let $y=A~cos(x)+B~sin(x)$ the differential equation by eliminating A and B constants is given by:

• A. $\frac{d^{2}y}{dx^{2}}-y^{2}=0$
• B. $\frac{d^{2}y}{dx^{2}}+y^{2}=0$
• C. $\frac{d^{2}y}{dx^{2}}-y=0$
• D. $\frac{d^{2}y}{dx^{2}}+y=0$

Hint:

For solutions involving $sin(ax)$ and $cos(ax)$, the resulting differential equation is always $y'' + a^2y = 0$. Here $a=1$.

Answer 1

The Correct Option is D

Concept: To form a differential equation from a general solution, differentiate the function with respect to $x$ as many times as there are independent arbitrary constants (in this case, two: A and B).

Step 1: First differentiation.
$y = A \cdot cos(x) + B \cdot sin(x)$ $\frac{dy}{dx} = -A \cdot sin(x) + B \cdot cos(x)$

Step 2: Second differentiation.
$\frac{d^2y}{dx^2} = -A \cdot cos(x) - B \cdot sin(x)$

Step 3: Eliminate constants and simplify.
Factor out the negative sign: $\frac{d^2y}{dx^2} = -(A \cdot cos(x) + B \cdot sin(x))$ Observe that the term inside the parentheses is the original function $y$. $\frac{d^2y}{dx^2} = -y$ $\frac{d^2y}{dx^2} + y = 0$. This is a standard linear homogeneous second-order differential equation.