# Recent questions and answers in Differential Equations

If the Laplace transform of a function $f(t)$ is given by $\frac{s+3}{\left ( s+1 \right )\left ( s+2 \right )}$, then $f(0)$ is $0$ $\frac{1}{2}$ $1$ $\frac{3}{2}$
Consider the following differential equation $\left ( 1+y \right )\frac{dy}{dx}=y.$ The solution of the equation that satisfies condition $y(1)=1$ is $2ye^{y}=e^{x}+e$ $y^{2}e^{y}=e^{x}$ $ye^{y}=e^{x}$ $\left ( 1+y \right )e^{y}=2e^{x}$
If $y(x)$ satisfies the differential equation $(\sin x) \dfrac{\mathrm{d}y }{\mathrm{d} x} + y \cos x = 1,$ subject to the condition $y(\pi/2) = \pi/2,$ then $y(\pi/6)$ is $0$ $\frac{\pi}{6}$ $\frac{\pi}{3}$ $\frac{\pi}{2}$
The Dirac-delta function $\left ( \delta \left ( t-t_{0} \right ) \right )$ for $\text{t}$, $t_{0} \in \mathbb{R}$ ... $\mathcal{L}\left ( \delta \left ( t-a \right ) \right )=F\left ( s \right )$ is $0$ $\infty$ $e^{sa}$ $e^{-sa}$
The ordinary differential equation $\dfrac{dy}{dt}=-\pi y$ subject to an initial condition $y\left ( 0 \right )=1$ is solved numerically using the following scheme: $\frac{y\left ( t_{n+1} \right )-y\left ( t_{n} \right )}{h}=-\pi y\left ( t_{n} \right )$ where $\text{h}$ is the ... $h$ in the interval ___________________. $0< h< \frac{2}{\pi }$ $0< h< 1$ $0< h< \frac{\pi }{2}$ for all $h> 0$
The solution of $\dfrac{d^2y}{dt^2}-y=1,$ which additionally satisfies $y \bigg \vert_{t=0} = \dfrac{dy}{dt} \bigg \vert_{t=0}=0$ in the Laplace $s$-domain is $\dfrac{1}{s(s+1)(s-1)} \\$ $\dfrac{1}{s(s+1)} \\$ $\dfrac{1}{s(s-1)} \\$ $\dfrac{1}{s-1} \\$
The Laplace transform of a function $f(t)$ is $L( f )=\dfrac{1}{(s^{2}+\omega ^{2})}.$ Then, $f(t)$ is $f\left ( t \right )=\dfrac{1}{\omega ^{2}}\left ( 1-\cos\:\omega t \right ) \\$ $f\left ( t \right )=\dfrac{1}{\omega}\cos\:\omega t \\$ $f\left ( t \right )=\dfrac{1}{\omega}\sin\:\omega t \\$ $f\left ( t \right )=\dfrac{1}{\omega^{2}}\left ( 1-\sin\:\omega t \right )$
For the equation $\dfrac{dy}{dx}+7x^2y=0$, if $y(0)=3/7$, then the value of $y(1)$ is $\dfrac{7}{3}e^{-7/3} \\$ $\dfrac{7}{3}e^{-3/7} \\$ $\dfrac{3}{7}e^{-7/3} \\$ $\dfrac{3}{7}e^{-3/7}$
The differential equation $\dfrac{dy}{dx}+4y=5$ is valid in the domain $0 \leq x \leq 1$ with $y(0)=2.25$. The solution of the differential equation is $y=e^{-4x}+5$ $y=e^{-4x}+1.25$ $y=e^{4x}+5$ $y=e^{4x}+1.25$
A harmonic function is analytic if it satisfies the Laplace equation. If $u(x,y)=2x^2-2y^2+4xy$ is a harmonic function, then its conjugate harmonic function $v(x,y)$ is $4xy-2x^2+2y^2+ \text{constant}$ $4y^2-4xy + \text{constant}$ $2x^2-2y^2+ xy + \text{constant}$ $-4xy+2y^2-2x^2+ \text{constant}$
Consider the differential equation $x^2 \dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-4y=0$ with the boundary conditions of $y(0)=0$ and $y(1)=1$. The complete solution of the differential equation is $x^2 \\$ $\sin \left (\dfrac{\pi x}{2} \right ) \\$ $e^x \sin \left (\dfrac{\pi x}{2} \right ) \\$ $e^{-x} \sin \left (\dfrac{\pi x}{2} \right) \\$
Given the ordinary differential equation $\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-6y=0$ with $y(0)=0$ and $\dfrac{dy}{dx}(0)=1$, the value of $y(1)$ is __________ (correct to two decimal places).
If $y$ is the solution of the differential equation $y^3 \dfrac{dy}{dx}+x^3 = 0, \: y(0)=1,$ the value of $y(-1)$ is $-2$ $-1$ $0$ $1$
Consider a function $u$ which depends on position $x$ and time $t$. The partial differential equation $\frac{\partial u}{\partial t} = \frac{\partial^2 u }{\partial x^2}$ is known as the Wave equation Heat equation Laplace's equation Elasticity equation
$F(s)$ is the Laplace transform of the function $f(t) =2t^2 e^{-t}$. $F(1)$ is _______ (correct to two decimal places).
An explicit forward Euler method is used to numerically integrate the differential equation $\dfrac{dy}{dt} = y$ using a time step of $0.1$. With the initial condition $y(0)=1$, the value of $y(1)$ computed by this method is ________ (correct to two decimal places)
Consider the differential equation $3y" (x)+27 y (x)=0$ with initial conditions $y(0)=0$ and $y'(0)=2000$. The value of $y$ at $x=1$ is ________.
The Laplace transform of $te^{t}$ is $\dfrac{s}{(s+1)^{2}} \\$ $\dfrac{1}{(s-1)^{2}} \\$ $\dfrac{1}{(s+1)^{2}} \\$ $\dfrac{s}{(s-1)}$
Consider the following partial differential equation for $u(x, y)$, with the constant $c > 1$: $\dfrac{\partial u}{\partial y}+c\dfrac{\partial u}{\partial x}=0$ Solution of this equation is $u(x, y) = f (x+cy)$ $u(x, y) = f (x-cy)$ $u(x, y) = f (cx+y)$ $u(x, y) = f (cx-y)$
The differential equation $\dfrac{d^{2}y}{dx^{2}}+16y=0$ for $y(x)$ with the two boundary conditions $\dfrac{dy}{dx}\bigg \vert _{x=0}=1$ and $\dfrac{dy}{dx}\bigg \vert_{x=\displaystyle \frac{\pi}{2}}=-1$ has. No solution. Exactly two solutions. Exactly one solution. Infinitely many solutions.
Solutions of Laplace’s equation having continuous second-order partial derivatives are called biharmonic functions harmonic functions conjugate harmonic functions error functions
Laplace transform of $\cos( \omega t)$ is $\dfrac{s}{s^2+\omega ^2} \\$ $\dfrac{\omega }{s^2+\omega ^2} \\$ $\dfrac{s}{s^2-\omega ^2} \\$ $\dfrac{\omega }{s^2-\omega ^2}$
If $f(t)$ is a function defined for all $t \geq 0$, its Laplace transform $F(s)$ is defined as $\int_{0}^{\infty }e^{st}f(t)dt \\$ $\int_{0}^{\infty }e^{-st}f(t)dt \\$ $\int_{0}^{\infty }e^{ist}f(t)dt \\$ $\int_{0}^{\infty }e^{-ist}f(t)dt$
Laplace transform of the function $f(t)$ is given by $F(s)=L\begin{bmatrix} f(t) \end{bmatrix}=\int_{0}^{\infty }f(t)e^{-st}dt$ . Laplace transform of the function shown below is given by $\displaystyle{\frac{1-e^{-2s}}{s}} \\$ $\displaystyle{\frac{1-e^{-s}}{2s}} \\$ $\displaystyle{\frac{2-2e^{-s}}{s}} \\$ $\displaystyle{\frac{1-2e^{-s}}{s}}$
Consider the following differential equation: $\dfrac{dy}{dt}=-5y$; initial condition: $y=2$ at $t=0$. The value of $y$ at $t=3$ is $-5e^{-10}$ $2e^{-10}$ $2e^{-15}$ $-15e^{2}$
The Laplace transform of $e^{i5t}$ where $i=\sqrt{-1}$, is $\dfrac{s-5i}{s^2-25} \\$ $\dfrac{s+5i}{s^2+25} \\$ $\dfrac{s+5i}{s^2-25} \\$ $\dfrac{s-5i}{s^2+25}$
Find the solution of $\dfrac{d^2y}{dx^2}=Y$ which passes through the origin and the point $\left(\ln 2,\dfrac{3}{4}\right)$ $y=\dfrac{1}{2}e^x-e^{-x}$ $y=\dfrac{1}{2}(e^x+e^{-x})$ $y=\dfrac{1}{2}(e^x-e^{-x})$ $y=\dfrac{1}{2}e^x+e^{-x}$
The Blasius equation related to boundary layer theory is a third-order linear partial differential equation third-order nonlinear partial differential equation second-order nonlinear ordinary differential equation third-order nonlinear ordinary differential equation
Consider the following statements regarding streamline(s): It is a continuous line such that the tangent at any point on it shows the velocity vector at that point There is no flow across streamlines $\dfrac{dx}{u}=\dfrac{dy}{v}=\dfrac{dz}{w}$ is the differential equation of a streamline, where $u$, $v$ and $w$ ... true? $(i), (ii), (iv)$ $(ii), (iii), (iv)$ $(i), (iii), (iv)$ $(i), (ii), (iii)$
Laplace transform of $\cos(\omega t)$ is $\dfrac{s}{s^2+\omega ^2}$. The Laplace transform of $e^{-2t} \cos(4t)$ is $\dfrac{s-2}{(s-2)^2+16} \\$ $\dfrac{s+2}{(s-2)^2+16} \\$ $\dfrac{s-2}{(s+2)^2+16} \\$ $\dfrac{s+2}{(s+2)^2+16}$
The solution of the initial value problem $\dfrac{dy}{dx}=-2xy$ ; $y(0)=2$ is $1+e^{{-x}^2}$ $2e^{{-x}^2}$ $1+e^{{x}^2}$ $2e^{{x}^2}$
Consider two solutions $x(t)=x_1(t)$ and $x(t)=x_2(t)$ of the differential equation $\dfrac{d^2x(t)}{dt^2}+x(t)=0, \: t>0$ such that $x_1(0)=1, \dfrac{dx_1(t)}{dt} \bigg \vert_{t=0}=0$, $x_2(0)=0, \dfrac{dx_2(t)}{dt}\bigg \vert _{t=0}=1$. The Wronskian $W(t)=\begin{bmatrix} x_1(t) & x_2(t)\\ \\ \dfrac{dx_1(t)}{dt} & \dfrac{dx_2(t)}{dt} \end{bmatrix}$ at $t=\pi /2$ is $1$ $-1$ $0$ $\pi /2$
The general solution of the differential equation $\dfrac{dy}{dx}=\cos(x+y)$, with $c$ as a constant, is $y+\sin \left (x+y \right )=x+c \\$ $\tan \left (\dfrac{x+y}{2} \right)=y+c \\$ $\cos \left (\dfrac{x+y}{2} \right )=x+c \\$ $\tan \left (\dfrac{x+y}{2} \right )=x+c$
The solution to the differential equation $\dfrac{d^2u}{dx^2}-k\dfrac{du}{dx}=0$ where $k$ is a constant, subjected to the boundary conditions $u(0)$ = $0$ and $u(L)$ = $U$, is $u=U\dfrac{x}{L}$ $u=U\left(\dfrac{1-e^{kx}}{1-e^{kL}}\right)$ $u=U\left(\dfrac{1-e^{-kx}}{1-e^{-kL}}\right)$ $u=U\left(\dfrac{1+e^{kx}}{1+e^{kL}}\right)$
The function $f(t)$ satisfies the differential equation $\dfrac{d^2f}{dt^2}+f=0$ and the auxiliary conditions, $f(0)=0$, $\dfrac{d(f)}{d(t)}(0)=4$. The Laplace transform of $f(t)$is given by $\dfrac{2}{s+1} \\$ $\dfrac{4}{s+1} \\$ $\dfrac{4}{s^2+1} \\$ $\dfrac{2}{s^4+1}$
The partial differential equation $\dfrac{\partial u }{\partial t}+u\dfrac{\partial u}{\partial x}=\dfrac{\partial^2 u}{\partial x^2}$ is a linear equation of order $2$ non-linear equation of order $1$ linear equation of order $1$ non-linear equation of order $2$