Question

If $\cos x$ and $\sin x$ are solutions of the different equation $a_0\frac{d^{2}y}{d{x}^2}+a_1\frac{dy}{dx}+a_{2}y=0$, where $a_0,a_1,a_2$ are real constants, then which of the following is/are always true?

• A. $A\cos x+B\sin x$ is a solution, where $A$ and $B$ are real costants
• B. $A\cos (x+\frac{\pi }{4})$ is a solution, where $A$ is real constant
• C. $A\cos x\sin x$ is a solution, where $A$ is real constant
• D. $A\cos (x+\frac{\pi }{4})+B\sin (x-\frac{\pi }{4})$ is a solution, where $A$ and $B$ are real constants

Answer 1

Answer: (A), (B), (D)

The correct options are
A $A\cos x+B\sin x$ is a solution, where $A$ and $B$ are real costants
B $A\cos (x+\frac{\pi }{4})$ is a solution, where $A$ is real constant
D $A\cos (x+\frac{\pi }{4})+B\sin (x-\frac{\pi }{4})$ is a solution, where $A$ and $B$ are real constants

Given differential equation is $a_0\frac{d^{2}y}{{dx}^2}+a_1\frac{dy}{dx}+a_{2}y=0$

Also given that, $\cos x$ and $\sin x$ are solutions to the given differential equation.

$⟹A\cos x+B\sin x$ is a solution to the given differential equation as it is homogenous, where $A$ and $B$ are real constants.

$A\cos (x+\frac{\Pi }{4})=\frac{A}{\sqrt{2}}(-\sin x+\cos x)=C_1\sin x+C_2\cos x$ is also a solution where $A=-B$

Consider, $A\cos (x+\frac{\Pi }{4})+B\sin (x-\frac{\Pi }{4})$

$=\frac{A}{\sqrt{2}}(\cos x-\sin x)+\frac{B}{\sqrt{2}}(\cos x-\sin x)$

$=(\frac{A}{\sqrt{2}}+\frac{B}{\sqrt{2}})\cos x-(\frac{A}{\sqrt{2}}+\frac{B}{\sqrt{2}})\sin x$

which is of the form$C_1\cos x+C_2\sin x$.

Hence, options A, B, and D are true.