Question

If $\cos x$ and $\sin x$ are solutions of the differential equation $a_0\frac{d^{2}y}{d{x}^2}+a_1\frac{dy}{dx}+a_{2}y=0$, where $a_0,a_1$ and $a_2$ are all 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 real constants
• 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 $\cos (x+\frac{\pi }{4})$ is a solution, where A is real constant
B A $\cos x+B\sin x$ is a solution, where A and B real constants
D A $\cos (x+\frac{\pi }{4})+B\sin (x-\frac{\pi }{4})$ is a solution, where A and B are real constants

(a) Let $f\left(x\right)=cosx$ and $g\left(x\right)=sinx$
Consider the Wronskian of $f\left(x\right)$ and $g\left(x\right)$,
$W=\left|\begin{array}{cc}f\left(x\right)& g\left(x\right)\\ f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\end{array}\right|$

$=\left|\begin{array}{cc}cosx& sinx\\ -sinx& cosx\end{array}\right|$

$=co{s}^{2}x+si{n}^{2}x=1\ne 0$

Thus, the functions are linearly independent. So, the general solution of given differential equation is given by $y=Acosx+Bsinx$, where A and B are real constants.

[$\because$ if $y_1$ and $y_2$ are linearly independent solutions of the differential equation $ay"+b{y}^{\prime }+c=0$, 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 $y=Acos(x+\frac{\pi }{4})$
$=A(cosx\cdot cos\frac{\pi }{4}-sinx\cdot sin\frac{\pi }{4})$
$=\frac{A}{\sqrt{2}}(cosx-sinx)$
$=\frac{A}{\sqrt{2}}cosx+(-\frac{A}{\sqrt{2}})sinx$
which is in the form of general solution.
Hence, option (b) is true.

(c) Let $y=Acosxsinx$, Which cannot be expressed in the form of general solution.

(d) Let $y=Acos(x+\frac{\pi }{4})+Bsin(x-\frac{\pi }{4})$
$=A(cosx\cdot \frac{1}{\sqrt{2}}-sinx\cdot \frac{1}{\sqrt{2}})+B(sinx\cdot \frac{1}{\sqrt{2}}-cosx\cdot \frac{1}{\sqrt{2}})$
$=cosx(\frac{A}{\sqrt{2}}-\frac{B}{\sqrt{2}})+sinx(\frac{B}{\sqrt{2}}-\frac{A}{\sqrt{2}})$ which is in the form of general solution.
Hence, option (d) is true.