# Differential Equations

 Question 1
For the exact differential equation, $\frac{du}{dx}=\frac{-xu^2}{2+x^2u}$, which one of the following is the solution ?
 A $u^2+2x^2=\text{Constant}$ B $xu^2+u=\text{Constant}$ C $\frac{1}{2}x^2 u^2+2u=\text{Constant}$ D $\frac{1}{2} u x^2+2x=\text{Constant}$
GATE ME 2022 SET-2   Engineering Mathematics
Question 1 Explanation:
Given D.E. is $\frac{du}{dx}=\frac{-xu^2}{2+x^2u}$
$\Rightarrow (2+x^2u)du+xu^2dx=0$
Here $M=xu^2 \;\;\;\;N=2+x^2u$
$\frac{\partial N}{\partial x}=2xu \;\;\;\;\;\frac{\partial M}{\partial x}=2xu$
This is exact D.E
General solution is
$\int_{\text{u constant}}Mdx+\int_{\text{term without 'x'}}Ndx=c$
$\Rightarrow \int (xu^2)dx+\int 2du=c$
$\therefore \; \frac{x^2u^2}{2}+2u=constant$
 Question 2
A polynomial $\phi (s)=a_{n}s^{n}+a_{n-1}s^{n-1}+...+a_{1}s+a_0$ of degree $n \gt 3$ with constant real coefficients $a_n, a_{n-1},...a_0$ has triple roots at $s=-\sigma$. Which one of the following conditions must be satisfied?
 A $\phi (s)=0$ at all the three values of s satisfying $s^3+\sigma ^3=0$ B $\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^2 \phi (s)}{ds^2}=0 \text{ at }s=-\sigma$ C $\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^4 \phi (s)}{ds^4}=0 \text{ at }s=-\sigma$ D $\phi (s)=0, \text{ and }\frac{d^3 \phi (s)}{ds^3}=0 \text{ at }s=-\sigma$
GATE ME 2022 SET-2   Engineering Mathematics
Question 2 Explanation:
Since $\varphi (s)$ has a triple roots at $s=-\sigma$
Therefore, $\varphi (s)=(s+\sigma )^3\psi (s)$
It satisfies all the conditions in option (B) is correct.
 Question 3
Solution of $\triangledown^2T=0$ in a square domain ($0 \lt x \lt 1$ and $0 \lt y \lt 1$) with boundary conditions:
$T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y$ is
 A $T(x,y)=x-xy+y$ B $T(x,y)=x+y$ C $T(x,y)=-x+y$ D $T(x,y)=x+xy+y$
GATE ME 2022 SET-1   Engineering Mathematics
Question 3 Explanation:
T(x, 0) = x $\Rightarrow$ option (c) is not correct.
T(0, y) = y $\Rightarrow$ all options satisfied.
T(x, 1) = 1 + x; $\Rightarrow$ only option (b) is satisfied.
T(1, y) = 1 + y is $\Rightarrow$ only option (b) is satisfied.
 Question 4
Consider the following differential equation
$(1+y)\frac{dy}{dx}=y$
The solution of the equation that satisfies the condition is $y(1)=1$ is
 A $2ye^y=e^x+e$ B $y^2e^y=e^x$ C $ye^y=e^x$ D $(1+y)e^y=2e^x$
GATE ME 2021 SET-2   Engineering Mathematics
Question 4 Explanation:
\begin{aligned} (1+y) \frac{d y}{d x} &=y \\ \Rightarrow\qquad \left(\frac{1}{y}+1\right) d y &=d x \\ \Rightarrow\qquad \log y+y &=x+c \\ \text { Using, } \qquad y(1) &=1 \\ \quad \log 1+1 &=1+c \quad \Rightarrow c=0 \\ \text { Hence, } \quad \log y+y &=x \\ \Rightarrow\qquad \log y+y \operatorname{loge} &=x \\ \log _{\mathrm{e}}\left(y \cdot e^{y}\right) &=x \\ \Rightarrow\qquad y e^{y} &=e^{x} \end{aligned}
 Question 5
If the Laplace transform of a function $f(t)$ is given by $\frac{s+3}{(s+1)(s+2)}$ , then $f(0)$ is
 A 0 B $\frac{1}{2}$ C 1 D $\frac{3}{2}$
GATE ME 2021 SET-2   Engineering Mathematics
Question 5 Explanation:
By using partial fraction concept.
\begin{aligned} f(t) &=L^{-1}\left[\frac{s+3}{(s+1)(s+2)}\right] \\ &=L^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right] \\ \Rightarrow \qquad f(t) &=2 e^{-t}-e^{-2 t} \\ \text { So, } \qquad f(c)&=2 e^{0}-e^{0}=2-1=1 \end{aligned}
 Question 6
The ordinary differential equation $\frac{dy}{dt}=-\pi y$ subject to an initial condition $y(0)=1$ is solved numerically using the following scheme:

$\frac{y(t_{n+1})-y(t_n)}{h}=-\pi y(t_n)$

where $h$ is the time step, $t_n=nh,$ and $n=0,1,2,...$. This numerical scheme is stable for all values of $h$ in the interval.
 A $0 \lt h \lt \frac{2}{\pi}$ B $0 \lt h \lt 1$ C $0 \lt h \lt \frac{\pi}{2}$ D for all $h \gt 0$
GATE ME 2021 SET-1   Engineering Mathematics
Question 6 Explanation:
\begin{aligned} \frac{y\left(t_{n+1}\right)-y\left(t_{n}\right)}{h} &=-\pi y\left(t_{n}\right) \\ y_{n+1} &=-\pi / y_{n}+y_{n}=(-\pi h+1) y_{n} \end{aligned}
It is recursion relation between $y_{n+1}$ and $y_{n}$
So solution will be stable if
\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}
Therefore option (A) is correct.
 Question 7
The Dirac-delta function $(\delta (t-t_0)) \text{ for }t,t_0 \in \mathbb{R},$ has the following property

$\int_{a}^{b}\varphi (t)\delta (t-t_0)dt=\left\{\begin{matrix} \varphi (t_0) & a \lt t_0 \lt b\\ 0 &\text{otherwise} \end{matrix}\right.$

The Laplace transform of the Dirac-delta function $\delta (t-a)$ for $a \gt 0$;
$\mathcal{L} (\delta (t-a))=F(s)$ is
 A 0 B $\infty$ C $e^{sa}$ D $e^{-sa}$
GATE ME 2021 SET-1   Engineering Mathematics
Question 7 Explanation:
\begin{aligned} \because \qquad \int_{0}^{-} f(t) \delta(t-a) d t&=f(a) \\ \therefore \qquad L\{\delta(t-a)\}&=\int_{0}^{-} e^{-s t} \delta(t-a) d t=e^{-a s} \end{aligned}
 Question 8
If $y(x)$ satisfies the differential equation

$(\sin x)\frac{dy}{dx}+y \cos x =1$

subject to the condition $y(\pi /2)=\pi /2$, then $y(\pi /6)$ is
 A 0 B $\frac{\pi}{6}$ C $\frac{\pi}{3}$ D $\frac{\pi}{2}$
GATE ME 2021 SET-1   Engineering Mathematics
Question 8 Explanation:
\begin{aligned} \frac{d y}{d x}+y \cot x&=\text{cosec} x\\ 1.F. \qquad&=e^{\int \cot x d x}=e^{\log \sin x}=\sin x\\ \Rightarrow \quad y(\sin x)&=\int \text{cosec} x \sin x d x+c\\ \Rightarrow \qquad y \sin x&=x+c\\ \Rightarrow \qquad \frac{\pi}{2} \sin \frac{\pi}{2} & =\frac{\pi}{2}+c \\ \Rightarrow \qquad \frac{\pi}{2} & =\frac{\pi}{2}+c \quad \Rightarrow c=0 \\ \Rightarrow \qquad y \sin x & =x\\ \Rightarrow \qquad y \sin \frac{\pi}{6}&=\frac{\pi}{6}\\ \Rightarrow \qquad y\left(\frac{1}{2}\right) &=\frac{\pi}{6} \\ \Rightarrow y &=\frac{\pi}{3} \end{aligned}
 Question 9
The solution of
$\frac{d^2y}{dt^2}-y=1,$
which additionally satisfies $y|_{t=0}=\left.\begin{matrix} \frac{dy}{dt} \end{matrix}\right|_{t=0}=0$ in the Laplace s-domain is
 A $\frac{1}{s(s+1)(s-1)}$ B $\frac{1}{s(s+1)}$ C $\frac{1}{s(s-1)}$ D $\frac{1}{s-1}$
GATE ME 2020 SET-2   Engineering Mathematics
Question 9 Explanation:
\begin{aligned} y^{\prime\prime}-y &=1 \\ y(0) &=1 \\ y^{\prime}(0) &=1 \\ L\left\{y^{\prime\prime}-y\right\} &=L\{1\} \\ s^{2} Y(s)-s y(0)-y^{\prime}(0)-y(s) &=\frac{1}{s} \\ y(s) &=\frac{1}{s\left(s^{2}-1\right)}\\ & =\frac{1}{s(s+1)(s-1)} \end{aligned}
 Question 10
The Laplace transform of a function f(t) is $L(f)=\frac{1}{s^2+\omega ^2}$. Then f(t) is
 A $f(t)=\frac{1}{\omega ^2}(1-\cos \omega t)$ B $f(t)=\frac{1}{\omega} \cos \omega t$ C $f(t)=\frac{1}{\omega} \sin \omega t$ D $f(t)=\frac{1}{\omega ^2}(1-\sin \omega t)$
GATE ME 2020 SET-1   Engineering Mathematics
Question 10 Explanation:
$L(t)=\frac{1}{s^{2}+\omega ^{2}}$
$f(t)=L^{-1}\left \{ \frac{1}{s^{2}+\omega ^{2}} \right \}=\frac{1}{\omega }\sin \omega t$
There are 10 questions to complete.