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 k is a constant, the general solution of \frac{d y}{d x}-\frac{y}{x}=1 will be in the form of
A
y=x ln(kx)
B
y=k ln(kx)
C
y=x ln(x)
D
y=xk ln(k)
GATE CE 2021 SET-2   Engineering Mathematics
Question 2 Explanation: 
\begin{aligned} \frac{d y}{d x}-\frac{y}{x} &=1 \\ \frac{d y}{d x}+P y &=Q \\ P &=-\frac{1}{x}, Q=1 \\ I F &=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=\frac{1}{x} \\ y(I F) &=\int Q(I F) d x+c \\ y\left(\frac{1}{x}\right) &=\int 1 \cdot \frac{1}{x} d x+\ln k \\ y &=x \ln (x k) \end{aligned}
Question 3
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 3 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 4
In the open interval \left ( 0,1 \right ), the polynomial p\left ( x \right) =x^{4}-4x^{3}+2 has
A
two real roots
B
one real root
C
three real roots
D
no real roots
GATE EE 2021   Engineering Mathematics
Question 4 Explanation: 
\begin{aligned} x^{4}+2 &=4 x^{3} \\ f_{1}(x) &=x^{4}+2 \\ f_{2}(x) &=4 x^{3} \end{aligned}
It is clear that point of intersection of these graphs is solution (or) root of p(x) = 0


According to intermediate value theorem
P(0) and P(1) are having opposite signs
\therefore a root of p(x) = 0 in (0, 1)
and also from graph, there is only one point of intersection
Hence exactly one real root exists in (0, 1)
Question 5
The solution of the second-order differential equation \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=0 with boundary conditions y(0)=1 and y(1)=3 is
A
e^{-x}+(3 e-1) x e^{-x}
B
e^{-x}-(3 e-1) x e^{-x}
C
e^{-x}+\left[3 e \sin \left(\frac{\pi x}{2}\right)-1\right] x e^{-x}
D
e^{-x}-\left[3 e \sin \left(\frac{\pi x}{2}\right)-1\right] x e^{-x}
GATE CE 2021 SET-1   Engineering Mathematics
Question 5 Explanation: 
Given \begin{aligned} \left(D^{2}+2 D+1\right) & y=0 \\ & y(0)=1 \\ & y(1)=3 \end{aligned}
Auxiliary equation in m^{2}+2 m+1=0
\begin{aligned} \Rightarrow \qquad \qquad \qquad \qquad \quad &m=-1,-1\\ \mathrm{CF}&=\left(C_{1}+C_{2} x\right) e^{-x}\\ \text { And } \qquad \qquad \qquad \quad \mathrm{PI}&=0\\ C_{1} \text { solution is } \qquad \quad u y &=C F+P I \\ y &=\left(C_{1}+C_{2} x\right) e^{-x} \\ \text { Using } y(0)=1, \qquad C_{1} &=1 \\ \text { Using } y(1)=3, \qquad C_{2} &=3 e-1 \\ \text { Hence by eq. (i), } \qquad y &=[1+(3 e-1) x] e^{-x} \\ y &=e^{-x}+(3 e-1) x e^{-x} \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
Let f\left ( x \right ) be a real-valued function such that {f}'\left ( x_{0} \right )=0 for some x _{0} \in\left ( 0,1 \right ), and {f}''\left ( x \right )> 0 for all x \in \left ( 0,1 \right ). Then f\left ( x \right ) has
A
no local minimum in (0,1)
B
one local maximum in (0,1)
C
exactly one local minimum in (0,1)
D
two distinct local minima in (0,1)
GATE EE 2021   Engineering Mathematics
Question 7 Explanation: 
x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
Question 8
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 8 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 9
Consider the differential equation given below.
\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}
The integrating factor of the differential equation is
A
\left ( 1-x^{2} \right )^{-3/4}
B
\left ( 1-x^{2} \right )^{-1/4}
C
\left ( 1-x^{2} \right )^{-3/2}
D
\left ( 1-x^{2} \right )^{-1/2}
GATE EC 2021   Engineering Mathematics
Question 9 Explanation: 
\begin{aligned} \frac{d y}{d x}+\frac{x}{1-x^{2}} y&=x \sqrt{y}, \quad \text { IF }=?\\ \text{Divided by }\sqrt{y}\\ \frac{1}{\sqrt{y}} \frac{d y}{d x}+\frac{x}{1-x^{2}} \sqrt{y}&=x \\ 2 \frac{d u}{d x}+\frac{x}{1-x^{2}} u&=x\\ \text{Let }\qquad x \sqrt{y}&=u\\ \frac{1}{2 \sqrt{v}} \frac{d y}{d x}&=\frac{d u}{d x}\\ \Rightarrow \qquad \frac{d u}{d x}+\frac{x}{2\left(1-x^{2}\right)} u&=\frac{x}{2} \rightarrow \text{ lines diff. equ.} \\ \text { I. } F&=e^{\int \frac{x}{2\left(1-x^{2}\right)} d x}=e^{-\frac{1}{4} \log \left(1-x^{2}\right)}&=e^{\log \left(1-x^{2}\right) \frac{-1}{4}} \\ \text { I.F }&=\frac{1}{\left(1-x^{2}\right)^{\frac{1}{4}}} \end{aligned}
Question 10
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 10 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}


There are 10 questions to complete.

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