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.

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