Question

Find the differential equation of the family of curves, \[x=A\cos nt+B\sin nt\] where A and B are arbitrary constants.

Answer 1

Hint: In this problem, we have to find the differential equation of the family of curves, whose given equation is \[x=A\cos nt+B\sin nt\], where A and B are arbitrary constants. We can see that we have two arbitrary constants, so we can find the differential equation of the second order. We can then simplify the steps to get a differential equation for the given equation.

Complete step by step answer:
Here we have to find the differential equation of the family of curves.
We know that the given equation is,
\[\Rightarrow x=A\cos nt+B\sin nt\]……… (1)
We can see that the equation contains two arbitrary constants, so we can find the differential equation of second order and simplify it to get the required solution.
We can now differentiate the given equation (1), which is the first order differentiation, we get
\[\Rightarrow \dfrac{dx}{dt}=-An\sin nt+Bn\cos nt\] ……… (2)
We can now differentiate the above equation (2), which is the second order differentiation, we get
\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-A{{n}^{2}}\cos nt-B{{n}^{2}}\sin nt\]
We can write the above step as,
\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-{{n}^{2}}\left( A\cos nt+B\sin nt \right)\]
We can now replace equation (1) in the above step, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-{{n}^{2}}x \\
 & \Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}+{{n}^{2}}x=0 \\
\end{align}\]
Therefore, the differential equation of the family of curves, \[x=A\cos nt+B\sin nt\] is\[\dfrac{{{d}^{2}}x}{d{{t}^{2}}}+{{n}^{2}}x=0\].

Note: Students make mistakes while differentiating the given equation, we should always remember the differentiation formula to differentiate the given equation. Here we have used the formula, as we differentiate \[\sin nt\] we get \[n\cos nt\] and as if we differentiate \[\cos nt\] we will get \[-n\sin nt\].