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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
$\text{If y}=\text{Ae}^{\text{mx}}+\text{Be}^{\text{nx}},\text{ show that }\frac{\text{d}^2\text{y}}{\text{dx}^2}-(\text{m}+\text{n})\frac{\text{dy}}{\text{dx}}+\text{mny}=0$
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Answer

 $\text{y}=\text{Ae}^\text{mx}+\text{Be}^{\text{nx}}\dots(1)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{Ame}^{\text{mx}}+\text{Bne}^{\text{nx}}\dots(2)$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{Am}^2\text{e}^{\text{mx}}+\text{Bn}^2\text{e}^{\text{nx}}\dots(3)$
$\text{L.H.S.}=\frac{\text{d}^2\text{y}}{\text{dx}^2}-(\text{m}+\text{n})\frac{\text{dy}}{\text{dx}}+\text{mny} $
$=(\text{Am}^2\text{e}^{\text{mx}}+\text{Bn}^2\text{e}^{\text{nx}})-(\text{m}+\text{n})(\text{Ame}^{\text{mx}}+\text{Bne}^{\text{nx}})+\text{mn}(\text{Ae}^{\text{mx}}+\text{Be}^{\text{nx}})$
$[\because \text{of }(1),(2),(3)]$$$
$=\text{Am}^2\text{e}^{\text{mx}}+\text{Bn}^2\text{e}^{\text{nx}}-\text{Am}^2\text{e}^{\text{mx}}-\text{Bmne}^{\text{nx}}-\text{Amne}^{\text{mx}}$
$-\text{Bn}^2\text{e}^{\text{nx}}+\text{Amne}^{\text{mx}}+\text{Bmne}^{\text{nx}}$
$=0$
$=\text{R.H.S.}$

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