If cos x and sin x are solutions of the diffe | vedclass

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(A, B, D) Let $f(x)=\cos x$ and $g(x)=\sin x$. Consider the Wronskian of $f(x)$ and $g(x)$.$W = \begin{vmatrix} f(x) & g(x) \ f'(x) & g'(x) \end{vmat

Vedclass · Accepted Answer

$A \cos \left(x+\frac{\pi}{4}\right)+B \sin \left(x -\frac{\pi}{4}\right)$ is a solution, where $A$ and $B$ are real constants

Vedclass · Answer

(A, B, D) Let $f(x)=\cos x$ and $g(x)=\sin x$. Consider the Wronskian of $f(x)$ and $g(x)$.$W = \begin{vmatrix} f(x) & g(x) \ f'(x) & g'(x) \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1 
eq 0$.Since the Wronskian is non-zero, the functions are linearly independent. The general solution is $y = A \cos x + B \sin x$.$(a)$ $A \cos x + B \sin x$ is the general solution, so it is true.$(b)$ $A \cos(x + \frac{\pi}{4}) = A(\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}) = \frac{A}{\sqrt{2}} \cos x - \frac{A}{\sqrt{2}} \sin x$. This is of the form $C_1 \cos x + C_2 \sin x$, so it is true.$(c)$ $A \cos x \sin x = \frac{A}{2} \sin(2x)$, which is not of the form $A \cos x + B \sin x$, so it is false.$(d)$ $A \cos(x + \frac{\pi}{4}) + B \sin(x - \frac{\pi}{4}) = A(\frac{\cos x - \sin x}{\sqrt{2}}) + B(\frac{\sin x - \cos x}{\sqrt{2}}) = \cos x(\frac{A-B}{\sqrt{2}}) + \sin x(\frac{B-A}{\sqrt{2}})$. This is of the form $C_1 \cos x + C_2 \sin x$, so it is true.

If $\cos x$ and $\sin x$ are solutions of the differential equation $a_{0} \frac{d^{2} y}{d x^{2}}+a_{1} \frac{d y}{d x}+a_{2} y=0$ where $a_{0}, a_{1}$ and $a_{2}$ are real constants, then which of the following is/are always true?

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