MCQ
જો $y = a{e^{mx}} + b{e^{ - mx}}$, તો ${{{d^2}y} \over {d{x^2}}} - {m^2}y = $
- A${{m}^{2}}(a{{e}^{mx}}-b{{e}^{-mx}})$
- B$1$
- ✓$0$
- Dએકપણ નહીં
$\therefore \frac{{dy}}{{dx}} = am{e^{mx}} - mb{e^{ - mx}}$
Again $\frac{{{d^2}y}}{{d{x^2}}} = a{m^2}{e^{mx}} + {m^2}b{e^{ - mx}}$
==> $\frac{{{d^2}y}}{{d{x^2}}} = {m^2}(a{e^{mx}} + b{e^{ - mx}}) \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = {m^2}y$
or $\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.