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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.
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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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