Download App

Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
Choose the correct answer from the given four option.The differential equation for $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x},$ where $A$ and $B$ are arbitrary constants is:
  • A
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha^2\text{y}=0$
  • ✓
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
  • C
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha\text{y}=0$
  • D
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha\text{y}=0$
✓

Answer

Correct option: B.
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
Given, $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=-\alpha\text{A}\sin\alpha\text{x}+\alpha\text{B}\cos\alpha\text{x}$
Again, differentiating both sides $\text{w.r.t. x,}$ we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{A}\alpha^2\cos\alpha\text{x}-\alpha^2\text{B}\sin\alpha\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\alpha^2(\text{A}\cos\alpha\text{x}-\text{B}\sin\alpha\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\alpha^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$

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

More "M.C.Q (1 Marks)" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If  $y = \sqrt {\sec x + \sqrt {\sec x + \sqrt {\sec x + ......\infty } } } \,,$ then value of  $\int\limits_0^{\pi /3} {\left( {(2y - 1)\frac{{dy}}{{dx}}} \right)} \,dx$ is equal to $(\sec x > 0)$ -
→
The area bounded by curve $x + y = 2, x-$ axis, and $y$ axis is
→
$|a \times i{|^2} + |a \times j{|^2} + |a \times k{|^2} = $
→
If $y = a{x^{n + 1}} + b{x^{ - n}},$ then ${x^2}\frac{{{d^2}y}}{{d{x^2}}}$ equals to
→
If $y = a{e^x} + b{e^{ - x}} + c$ where $a,b,c$ are parameters then $y''' = $
→
Choose the correct answers from the given four options : For the function $\text{f(x)}=\text{x}+\frac{1}{\text{x}},\text{x}\in[1,3],$ the value of $c$ for mean value theorem is :
→
The solution of the differential equartion $\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$ is given by :
→
A continuous and differentiable function $‘ f ‘$ satisfies the condition ,$\int\limits_0^x \, f (t) d t = f^2 (x) - 1$ for all real $‘ x ‘$. Then :
→
If the distance  $‘s’$  traveled by a particle in time $ t$ is $s = a\sin t + b\cos 2t$, then the acceleration at $ t = 0$ is
→
Let $\quad P_1=I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$,

$P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$

where $P _{ K }^{ T }$ denotes the transpose of the matrix $P _{ K }$. Then which of the following options is/are correct?

$(1)$ $X -30 I$ is an invertible matrix

$(2)$ The sum of diagonal entries of $X$ is 18

$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$

$(4)$ $X$ is a symmetric matrix

→