Q113 / 160 · WBJEE 2015 · Mathematics
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?
- Correct answerA$A \cos x+B \sin x$ is a solution, where $A$ and $B$ are feal constants
- Correct answerB$A \cos \left(x+\frac{\pi}{4}\right)$ is a solution, where $A$ is a real constant
- C$A \cos x \sin x$ is a solution, where $A$ is a real constant
- Correct answerD$A \cos \left(x+\frac{\pi}{4}\right)+B \sin \left(x -\frac{\pi}{4}\right)$ is a solution, where A and $B$ are real constants
Solution
Let $f(x)=\cos x$ and $g(x)=\sin x$ Consider the Wronskian of $f(x)$ and $g(x)$.
$$
\begin{aligned}
W &=\left|\begin{array}{ll}
f(x) & g(x) \\
f^{\prime}(x) & g^{\prime}(x)
\end{array}\right| \\
&=\left|\begin{array}{ll}
\cos x & \sin x \\
-\sin x & \cos x
\end{array}\right| \\
&=\cos ^{2} x+\sin ^{2} x=1 \neq 0
\end{aligned}
$$
Thus, the functions are linearly independent. So, the general solution of given differential equation is given by $y=A \cos x+B \sin x,$ where $A$ and $B$ are real constants.
[$\because$ if $y_{1}$ and $y_{2}$ are linearly independent solutions of the differential equation $a y^{\prime \prime}+b y^{\prime}+c=0,$ then the general solution is $y=c_{1} y_{1}+c_{2} y_{2},$ where $c_{1}$ and $c_{2}$ are constants] Hence, option (a) is true.
(b) Let $y=A \cos \left(x+\frac{\pi}{4}\right)$
$$
\begin{array}{l}
=A\left(\cos x \cdot \cos \frac{\pi}{4}-\sin x \cdot \sin \frac{\pi}{4}\right) \\
=\frac{A}{\sqrt{2}}(\cos x-\sin x) \\
=\frac{A}{\sqrt{2}} \cos x+\left(-\frac{A}{\sqrt{2}}\right) \sin x
\end{array}
$$
which is in the form of general solution.
Hence, option (b) is true.
(c) Let $y=A \cos x \sin x$
which cannot be
expressed in the form of general solution.
(d) Let $y=A \cos \left(x+\frac{\pi}{4}\right)+B \sin \left(x-\frac{\pi}{4}\right)$
$$
\begin{array}{l}
=A\left(\cos x \cdot \frac{1}{\sqrt{2}}-\sin x \cdot \frac{1}{\sqrt{2}}\right) \\
\quad+B\left(\sin x \cdot \frac{1}{\sqrt{2}}-\cos x \cdot \frac{1}{\sqrt{2}}\right) \\
=\cos x\left(\frac{A}{\sqrt{2}}-\frac{B}{\sqrt{2}}\right)+\sin x\left(\frac{B}{\sqrt{2}}-\frac{A}{\sqrt{2}}\right) \text { which is }
\end{array}
$$
in the form of general solution. Hence, option (d) is true.
Hence, the correct answer is options: 'A', 'B', 'D'.