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Find the Differential Equation of the Family of Curves, X = a Cos Nt + B Sin Nt, Where a and B Are Arbitrary Constants. - Mathematics

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प्रश्न

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

योग
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उत्तर

The equation of the family of curves is \[x = A\cos nt + B\sin nt\]                              .........(1)

where `A" and "B` are arbitrary constants.

This equation contains two arbitrary constants, so we shall get a differential equation of second order.

Differentiating equation (1) with respect to t, we get

\[\frac{dx}{dt} = - \text{ An }\sin\text{ nt }+ \text{ Bn }\cos nt\]                          ............(2)

Differentiating equation (2) with respect to t, we get

\[\frac{d^2 x}{d t^2} = - A n^2 \cos nt - B n^2 \sin nt\]

\[ \Rightarrow \frac{d^2 x}{d t^2} = - n^2 \left( A\cos nt + B\sin nt \right)\]

\[ \Rightarrow \frac{d^2 x}{d t^2} = - n^2 x\]

\[ \Rightarrow \frac{d^2 x}{d t^2} + n^2 x = 0 \]

It is the required differential equation .

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अध्याय 22: Differential Equations - Exercise 22.02 [पृष्ठ १६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.02 | Q 5 | पृष्ठ १६

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