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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
If $\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$ and $\frac{\text{d}^2\text{x}}{\text{dt}^2}=\lambda\text{x}$ then find the value of $\lambda.$
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Answer

we have $\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{a}\sin(\text{nt})\times\text{n}-\text{bn}\cos)\text{nt}$$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{an}^2\cos(\text{nt})+\text{bn}^2\sin(\text{nt})$
since, $\frac{\text{d}^2\text{y}}{\text{dt}^2}=\lambda\text{x}$
$\Rightarrow-\text{an}^2\cos(\text{nt})+\text{bn}^2\sin(\text{nt})=\lambda(\text{a}\cos\text{nt}-\text{b}\sin\text{nt})$
$\Rightarrow\lambda=\text{n}^2$

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