Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
Question
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:
  1. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha^2\text{y}=0$
  2. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
  3. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha\text{y}=0$
  4. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha\text{y}=0$
✓

Answer

  1. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$

Solution:

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 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 the force $\overrightarrow F = i + 2j + 3k$ moves from $i + j - k$ to $2i - j + k,$ then work done will be represented by
→
The distance travelled by a particle moving in a striaght line in time $t$ is $s = \sqrt {a{t^2} + bt + c} $. Acceleration of the particle is
→
If A is a square matrix of order 3 and |A| = 5, then the value of |2A'| is:
  1. -10
  2. 10
  3. -40
  4. 40
→
If $\int\limits_0^{f(x)} {{t^2}\,dt} $ $= x\, \cos\, \pi\, x $, then $f ‘ (9)$
→
If $\int {{x^5}{e^{ - 4{x^3}}}\,dx = \frac{1}{{48}}{e^{ - 4{x^3}}}f\left( x \right) + C} $, where $C$ is a constant  of integration, then $f(x)$ is equal to
→
If $\int {\,\,\frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} $ $dx$  =$  A $ $ln$ $ |x| $ $+ $$\frac{B}{{1 + {x^2}}}$$ +$ $ c$ , where $c$  is the constant of integration then :
→
The area bounded by the parabola $x^2=4 y$ and its latus rectum is:
→
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(1 + \sin2\text{x})}}\text{dx}$ is:
  1. $\sin\text{x + c}$
  2. $\text{x + c}$
  3. $\cos\text{x + c}$
  4. $\tan\text{x + c}$
→
Let $A$ be a square matrix of order $2$ such that $|A|=2$ and the sum of its diagonal elements is $-3$ . If the points $(x, y)$ satisfying $A^2+x A+y I=0$ lie on a hyperbola, whose transverse axis is parallel to the x-axis, eccentricity is e and the length of the latus rectum is $\ell$, then $\mathrm{e}^4+\ell^4$ is equal to........................... 
→
Consider a rectangle whose length is increasing at the uniform rate of $2\, m/sec$, breadth is decreasing at the uniform rate of $3\, m/sec$ and the area is decreasing at the uniform rate of $5\,m^2/ sec$ . If after some time the breadth of the rectangle is $2\, m$ then the length of the rectangle is ........ $m.$
→