If y = a e^ mx + b e^ - mx , then d^2 y d x^2 - m^2 y = - Vidyadip

Question

If $y = a{e^{mx}} + b{e^{ - mx}}$, then ${{{d^2}y} \over {d{x^2}}} - {m^2}y = $

✓

Answer

c
(c) $y = a{e^{mx}} + b{e^{ - mx}};$

$\therefore \frac{{dy}}{{dx}} = am{e^{mx}} - mb{e^{ - mx}}$

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 $\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Suppose we have an arithmetic progression $a_1, a_2, \ldots a_n, \ldots$ with $a_1=1, a_2-a_1=5$. The median of the finite sequence $a_1, a_2, \ldots, a_k$, where $a_k \leq 2021$ and $a_{k+1} > 2021$ is

Solve $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2},$ then $x$ is equal to

Let $A _{1}, A _{2}, A _{3}, \ldots \ldots$ be an increasing geometric progression of positive real numbers. If $A _{1} A _{3} A _{5} A _{7}=\frac{1}{1296}$ and $A _{2}+ A _{4}=\frac{7}{36}$, then, the value of $A _{6}+ A _{8}+ A _{10}$ is equal to

The value of ${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}$ is

The differential equation ${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} - {\left( {\frac{{dy}}{{dx}}} \right)^{1/2}} = {y^3}$ has the degree

If $ a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d) = $

Consider a function $f : N \rightarrow R$, satisfying $f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x) ; x \geq 2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

If $\sin x=-\frac{3}{5}$, where $\pi < x < \frac{3 \pi}{2}$ then $80\left(\tan ^2 x-\cos x\right)$ is equal to :

Three points $O (0,0), P \left( a , a ^2\right), Q \left(- b , b ^2\right), a >0, b>0$, are on the parabola $y=x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola, and $S _2$ be the area of the triangle $\text{OPQ}$ . If the minimum value of $\frac{ S _1}{S_2}$ is $\frac{ m }{ n }, \operatorname{gcd}( m , n )=1$, then $m + n$ is equal to :

The area of the region bounded by $y = \,\,|x - 1|$ and $y = 1$ is