# Differential Equations

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 Explanation:
$L(t)=\frac{1}{s^{2}+\omega ^{2}}$
$f(t)=L^{-1}\left \{ \frac{1}{s^{2}+\omega ^{2}} \right \}=\frac{1}{\omega }\sin \omega t$
 Question 8
A differential equation is given as
$x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=4$
The solution of the differential equation in terms of arbitrary constants $C_1 \; and \; C_2$ is
 A $y=C_1 x^2+C_2 x+2$ B $y=\frac{C_1}{x^2}+C_2x+2$ C $y=C_1 x^2+C_2 x+2$ D $y=\frac{C_1}{x^2}+C_2x+4$
GATE ME 2019 SET-2   Engineering Mathematics
Question 8 Explanation:
Given $\left(x^{2} D^{2}-2 x D+2\right) y=4\dots(1)$
where $D=\frac{d}{d x}$
Let $x=e^{z}$ or $\log x=z \& x D=\theta, x^{2} D^{2}=\theta\left(\theta^{2}-1\right)\dots(2)$
Where $\theta=\frac{d}{d z}$
Using (2),(1) becomes
$\begin{array}{l} {[\theta(\theta-1)-2 \theta+2] y=4} \\ \Rightarrow\left(\theta^{2}-3 \theta+2\right) y=4 \\ \Rightarrow f(\theta) y=Q(z), \\ \text { where } f(\theta)=\theta^{2}-3 \theta+2 \& Q(z)=4 \end{array}$
C.F
Auxiliary equation if f(m)=0
$\begin{array}{l} \Rightarrow \mathrm{m}^{2}-3 \mathrm{m}+2=4 \\ \Rightarrow \mathrm{m}=1,2 \\ \Rightarrow \mathrm{y}_{\mathrm{C}}=\mathrm{C}_{1} \mathrm{e}^{2 \mathrm{z}}+\mathrm{C}_{2} \mathrm{e}^{\mathrm{z}} \\ \quad \mathrm{y}_{\mathrm{C}}=\mathrm{C}_{1} \mathrm{x}^{2}+\mathrm{C}_{2} \mathrm{x} \end{array}$
P.I:
$\theta(z)=4=4 \cdot e^{0 z}=k e^{a z+b}$
Here, $f(\theta)=f(a)=f(0)=0-0+2=2$
$\therefore y_{p}=\frac{1}{\int(a)} \theta(z)=\frac{1}{2} 4=2$
Hence, the solution of (1) is $y=y_{C}+y_{p}$
i.e., $y=\left(C_{1} x^{2}+C_{2} x\right)+2$
 Question 9
The differential equation $\frac{dy}{dx}+4y=5$ is valid in the domain $0\leq x\leq 1$ with y(0)=2.25. The solution of the differential equation is
 A $y=e^{-4x}+5$ B $y=e^{-4x}+1.25$ C $y=e^{4x}+5$ D $y=e^{4x}+1.25$
GATE ME 2019 SET-2   Engineering Mathematics
Question 9 Explanation:
$\begin{array}{l} \text { Given } \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}=5,0 \leq \mathrm{x} \leq 1 \; \; \ldots(1) \\ \because \frac{d y}{d x}+P(x, y)=Q(x) \\ \text { With } \mathrm{y}(0)=2.25\;\;\; \ldots(2) \\ \text { Here, I.F }=\mathrm{e}^{\int 4 \mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}} \end{array}$
The general solution of (1) is given by
$y \cdot e^{4 x}=\int(5)\left(e^{4 x}\right) d x+c$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{4 \mathrm{x}}=\frac{5}{4} \mathrm{e}^{4 \mathrm{x}}+\mathrm{c}\cdots(3)$
Using (2) and (3)
$(2.25)(1)=\left(\frac{5}{4}\right)(1)+c$
$c=1 \cdots(4)$
The solution of (1) from (3) and (4) is
$y \cdot e^{4 x}=\frac{5}{4} e^{4 x}+1$
or $y=\frac{5}{4}+e^{4 x}=e^{-4 x}+1.25$
 Question 10
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
 A $4xy-2x^2+2y^2+$ constant B $4y^2-4xy+$ constsnt C $2x^2-2y^2+xy+$ constsnt D $-4xy+2y^2-2x^2+$ constant
GATE ME 2019 SET-1   Engineering Mathematics
Question 10 Explanation:
\begin{aligned} &\text { Given } \mathrm{u}(\mathrm{x}, \mathrm{y})=2 \mathrm{x}^{2}-2 \mathrm{y}^{2}+4 \mathrm{xy}\\ &\frac{\partial u}{\partial x}=4 x+4 y, \frac{\partial u}{\partial y}=-4 y+4 x\\ &\text { By C-R equations } \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \& \frac{\partial u}{\partial y}=\frac{-\partial v}{\partial x}\\ &\therefore \frac{\partial v}{\partial x}=\frac{-\partial u}{\partial y}=4 y-4 x \; \; \ldots(1)\\ &\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=4 x+4 y \; \; \ldots(2)\\ &\begin{array}{l} \text { From }(1),\\ V=\int(4 y-4 x) d x+k(y) \\ V=4 x y-2 x^{2}+K(y) \;\; \ldots(3) \end{array}\\ &\therefore \frac{\partial v}{\partial y}=4 x+\frac{d}{d y} k(y)\ldots(4)\\ &\text { From }(2) \&(3), \\ &\frac{\mathrm{d}}{\mathrm{dy}} \mathrm{k}(\mathrm{y})=4 \mathrm{y}\\ &\int \frac{\mathrm{d}}{\mathrm{dy}} \mathrm{k}(\mathrm{y}) \mathrm{d} \mathrm{y}=\int 4 \mathrm{ydy}\\ &\mathrm{K}(\mathrm{y})=2 \mathrm{y}^{2}+\mathrm{C} \end{aligned}
while C is arbitrary real constant.
$\therefore$ Conjugate harmonic of u is
$\mathrm{V}=4 \mathrm{xy}-2 \mathrm{x}^{2}+2 \mathrm{y}^{2}+\mathrm{C}$
There are 10 questions to complete.