Question

Show that \[y = A\cos nx + B\sin nx\], is a solution of the differential equation: $\dfrac{{{d^2}y}}{{d{x^2}}} + {n^2}y = 0$.

Answer 1

Hint:
We see that the given differential equation is the second order differential equation. It means that we differentiate the equation \[y = A\cos nx + B\sin nx\] two times to obtain the second order differential equations. The differentiation is a method of finding a function that generates the rate of change between one variable and another variable. In this equation \[y = A\cos nx + B\sin nx\], y is the dependent variable and x is the independent variable.

Complete step by step solution:
The equation given in the problem is as follows.
\[y = A\cos nx + B\sin nx\].
We can differentiate the above equation with respect to x by using Chain rule.
$\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {A\cos nx + B\sin nx} \right)\\
\dfrac{{dy}}{{dx}} = - A\sin nx\left( {\dfrac{d}{{dx}}\left( {nx} \right)} \right) + B\cos nx\left( {\dfrac{d}{{dx}}nx} \right)\\
 = - nA\sin nx + Bn\sin nx$
The required differential equation needs the second order of differential equation therefore, we will again differentiate the above equation with respect to x.
$\dfrac{{{d^2}y}}{{d{x^2}}} = - An\cos nx\left( {\dfrac{d}{{dx}}\left( {nx} \right)} \right) + Bn\sin nx\left( {\dfrac{d}{{dx}}nx} \right)\\
 = - {n^2}A\cos nx - B{n^2}\sin nx\\
 = - {n^2}\left( {A\cos nx + B\sin nx} \right)$
We know that \[y = A\cos nx + B\sin nx\] which can be used in the above equation. So, substitute the value of \[\left( {A\cos nx + B\sin nx} \right)\] with y in the above expression.
$\dfrac{{{d^2}y}}{{d{x^2}}} = - {n^2}\left( y \right)\\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} + {n^2}y = 0$

Hence, the above result proves the required equation is \[\dfrac{{{d^2}y}}{{d{x^2}}} + {n^2}y = 0\]. Thus, it is proved that \[y = A\cos nx + B\sin nx\] is a solution of the differential equation: $\dfrac{{{d^2}y}}{{d{x^2}}} + {n^2}y = 0$.

Additional Information:
A differential equation is like an equation of dependent terms differentiated to the different orders. There are a lot of ways of solving such types of equations.

Note:
Make sure to use proper chain rules while doing the differentiation. You should know the basic formula of the trigonometry such as the differential of \[\sin x\] is \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] and the differential of \[\cos x\] is \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]. Also don’t get confused with the process of differentiation and the process of the integration.