Mathematics · Physics

Differential Equations

247 Questions

Differential equations involve finding functions that relate variables to their rates of change. These questions cover first and second order equations and separation of variables. They are a core part of the mathematics syllabus for high level competitive exams.

First order separationSecond order linear equationsInitial value problemsPredator prey modelsCauchy Riemann equations

Differential Equations Questions

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

The solution of the differential equation, $y\,dx + \left( {x + {x^2}y} \right)dy = 0$ is

  1. $\log y = cx$
  2. $ \log y- \dfrac{1}{{xy}} = c$
  3. $\dfrac{1}{{xy}} - \log y = c$
  4. $\dfrac{1}{{xy}} + \log y = c$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

$ydx+\left( { x+{ x^{ 2 } }y } \right) dy=0 \ ydx=-\left( { x+{ x^{ 2 } }y } \right) dy \ ydx+xdy=-{ x^{ 2 } }ydy \ \dfrac { { ydx+xdy } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y }  \ \dfrac { { d\left( { xy } \right)  } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y } $

On taking integrating both sides

$\begin{array}{l} \dfrac { { -1 } }{ { xy } } =-\log  y+c \\ \log  y-\dfrac { 1 }{ { xy } } =c \end{array}$


Hence, this is the answer.

Multiple choice physics option b: engineering physics introduction to sound free, forced and damped oscillations resonance

If we wish to represent the equation for the position of the mass in terms of a differential equation, which one of these would be the most suitable?

  1. $ m \dfrac{d^2x}{dt^2} + b \dfrac{dx}{dt} + kx = 0$
  2. $ m \dfrac{d^2x}{dt^2} - b \dfrac{dx}{dt} + kx = 0$
  3. $ m \dfrac{d^2x}{dt^2} + b \dfrac{dx}{dt} - kx = 0$
  4. $ m \dfrac{d^2x}{dt^2} -b \dfrac{dx}{dt} - kx = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force on body oscillating in resistive medium is 

$f = -kx - bv$
$\Rightarrow m\dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =-kx-b\dfrac { dx }{ dt } \ \Rightarrow m\dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +b\dfrac { dx }{ dt } +kx=0$
k = oscillating constant 
x = displacement of body from mean position 
b = constant depends on resistive medium 
v = velocity of object = $\dfrac{dx}{dt}$
m = mass of object .

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

General solution of the equation $ y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$ represents _____________.

  1. a straight line or hyperbola

  2. a straight line or parabola

  3. a parabola or hyperbola

  4. circles
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given differential equation can be rearranged into a form that describes a family of curves, which are hyperbolas or parabolas depending on the integration constants.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Differential equation of all hyperbolas which pass through the origin, and have their asymptotes parallel to the coordinate axes is?

  1. $xy\dfrac{d^2y}{dx^2}-2x\left(\dfrac{dy}{dx}\right)^2+2y=0$
  2. $xy\dfrac{d^2y}{dx^2}-2\left(\dfrac{dy}{dx}\right)^2+2y\left(\dfrac{dy}{dx}\right)=0$
  3. $xy\left(\dfrac{d^2y}{dx^2}\right)-2x\left(\dfrac{dy}{dx}\right)^2+2y\dfrac{dy}{dx}=0$
  4. $xy\dfrac{d^2y}{dx^2}+2x\left(\dfrac{dy}{dx}\right)^2+y\left(\dfrac{dy}{dx}\right)=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hyperbolas with asymptotes parallel to axes have the form (x-h)(y-k) = c. Differentiating twice leads to the differential equation xy(d^2y/dx^2) - 2x(dy/dx)^2 + 2y(dy/dx) = 0 (or similar depending on form). Option A is the standard differential equation for this family.

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

The general solution of the differential equation $\sin{2x}\left( \cfrac { dy }{ dx } -\sqrt { \tan { x }  }  \right) -y=0$ is $y\phi(x)=x+c$ then ${ \Phi  }^{ 1 }\left( \cfrac { \pi  }{ 4 }  \right) $ is _____

  1. $1$
  2. $-1$
  3. $\cfrac{1}{\sqrt{3}}$
  4. $\sqrt{-3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \frac{dy}{dx}-\text{cosec}2x y=\sqrt{\tan x};I.F=e^{\displaystyle \int-\text{cosec}2xdx}$

$=\sqrt{\text{cosec }2x+\text{cot} 2x} \implies y\sqrt{\text{cosec}2x+\text{cot}2x}=x+c\implies \phi'\bigg(\frac{\pi}{4}\bigg)=-1$

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

The second order differential equation is :

  1. ${ y' }^{ 2 }+x={ y }^{ 2 }$
  2. $y'+y''+y=\sin { x } $
  3. $y'''+y''+y=0$
  4. $y'=y$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that, the order of the differential equation is the order of highest derivative.
So, equation $y'+y''+y=\sin { x } $ is the second order differential equation.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

In a $\Delta ABC$ if sides a and b remain constant such that $\alpha$ is the error in C, then relative error in its area is

  1. $\alpha \cot C$
  2. $\alpha \sin C$
  3. $\alpha\tan C$
  4. $\alpha\cos C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Area of triangle $S\displaystyle =\dfrac {1}{2}ab \sin C$
$\displaystyle \Rightarrow \dfrac {dS}{dC}=\dfrac {1}{2}ab \cos C$
Now, approximate error in S is $\Delta S=\dfrac {dS}{dC}\Delta C$
$\displaystyle\Rightarrow \Delta S=\dfrac {1}{2}ab \cos C \alpha              [\because \Delta C=\alpha]$
$\displaystyle\Rightarrow \dfrac {\Delta S}{S}=\dfrac {\dfrac {1}{2}ab \cos C}{\dfrac {1}{2}ab \sin C}\alpha=\alpha \cot C$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

In a $\Delta ABC$ the sides b and c are given. If there is an error $\Delta A$ in measuring angle A, then the error $\Delta a$ in side a is given by

  1. $\dfrac {S}{2a}\Delta A$
  2. $\dfrac {2S}{a}\Delta A$
  3. bc sin A $\Delta A$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ABC$ we have
$\Rightarrow d\left( 2bc\cos { A }  \right) =d\left( { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } \right) \ \Rightarrow -2bc\sin { A } dA=-2ada\ \Rightarrow bc\sin { A } dA=ada$
$\displaystyle \Rightarrow \frac { 2 }{ a } \left( \frac { 1 }{ 2 } bc\sin { B }  \right) dA=da$
$\displaystyle \Rightarrow da=\frac { 2S }{ a } dA$
$\displaystyle \Rightarrow \triangle a=\frac { 2S }{ a } dA\ \left[ \because dx\equiv \triangle a\quad and\quad dA=BA \right] $

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If an error of $1^o$ is made in measuring the angle of a sector of radius $30 \ cm$, then the approximate error in its area is

  1. $450 cm^2$
  2. $25\pi cm^2$
  3. $2.5\pi cm^2$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Area of sector $\displaystyle A=\dfrac{\pi r^{2}\theta}{360}$
Given $r=30 cm, d{\theta}=1^{0}$
Approximate error in A is $\displaystyle=dA=(\dfrac{dA}{d\theta})\Delta \theta$
                             $\displaystyle = \dfrac{900\pi}{360} $
$\displaystyle \Rightarrow dA =2.5 \pi cm^{2}$
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A = \left[ {\begin{array}{*{20}{c}}1&2\3&4\end{array}} \right]$, then $8A^{-4}$ is equal to

  1. $145A^{-1}+27I$
  2. $145A^{-1}-27I$
  3. $27I - 145A^{-1}$
  4. $29A^{-1} +9I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the Cayley-Hamilton theorem for matrix A = [[1, 2], [3, 4]], the characteristic equation is A^2 - 5A - 2I = 0. By manipulating this equation, one can express higher powers of A in terms of A and I, eventually leading to the expression 27I - 145A^-1.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Latus rectum of the conic satisfying the differential equation $x dy+y dx=0$ and passing through the point $(2,8)$ is :

  1. $4\sqrt{2}$
  2. $8$
  3. $8\sqrt{2}$
  4. $16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The differential equation is

$xdy+ydx=0$
$d(xy)=0$
By integrating we get,
$xy=k$..........(1)

Equation (1) passes through the point $(2,8)$, so
$(2)(8)=k$
$k=16$

So, equation of the conic is 
$xy=16$
which is a rectangular hyperbola $(xy=c^{2})$, where $c=4$

Length of latus rectum for rectangular hyperbola is $2\sqrt{2}c=8\sqrt{2}$
 





Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Length of latusrectum of the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{b^{2}}=1$, if the normal, at an end of latusrectum passes through one extremity of the minor axis, then equation of eccentricity of ellipse is

  1. $e^4+e^2-1=0$
  2. <font face="MathJax_Main">$e^3+e^2-1=0$</font>
  3. $e^4+e^2+1=0$

  4. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><font face="MathJax_Main">none of these</font>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a specific derivation problem involving the normal at the end of the latus rectum. The resulting condition for eccentricity e is e^4 + e^2 - 1 = 0.