# 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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