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Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.2 question 5 maths

Answers (1)

Answer:

 \frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}}+n^{2}x=0

Hint:

Differentiating the given equation with respect to x.

Let the same equation and substitutes its value.

Given:

 x=Acos\: nt+Bsin\: nt

Solution:

x=Acos\: nt+Bsin\: nt

\begin{aligned} &\frac{\mathrm{d} x}{\mathrm{d} t}=-An\: sin(nt)+Bn\: cos(nt) \\ &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}= -An^{2}\: cos(nt)+Bn^{2}\: sin(nt) \\ &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}=-n^{2}[Acos\: (nt)+Bsin\: (nt)] \end{aligned}

We get,

\begin{aligned} &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}=-n^{2}x \\ &\frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}+n^{2}x=0 \end{aligned}

Posted by

Gurleen Kaur

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