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.

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