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
(1+y)\frac{dy}{dx}=y
The solution of the equation that satisfies the condition is y(1)=1 is
2ye^y=e^x+e | |
y^2e^y=e^x | |
ye^y=e^x | |
(1+y)e^y=2e^x |
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
y=x ln(kx) | |
y=k ln(kx) | |
y=x ln(x) | |
y=xk ln(k) |
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
0 | |
\frac{1}{2} | |
1 | |
\frac{3}{2} |
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}
\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
two real roots | |
one real root | |
three real roots | |
no real roots |
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)
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
e^{-x}+(3 e-1) x e^{-x} | |
e^{-x}-(3 e-1) x e^{-x} | |
e^{-x}+\left[3 e \sin \left(\frac{\pi x}{2}\right)-1\right] x e^{-x} | |
e^{-x}-\left[3 e \sin \left(\frac{\pi x}{2}\right)-1\right] x e^{-x} |
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}
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.
\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.
0 \lt h \lt \frac{2}{\pi} | |
0 \lt h \lt 1 | |
0 \lt h \lt \frac{\pi}{2} | |
for all h \gt 0 |
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.
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
no local minimum in (0,1) | |
one local maximum in (0,1) | |
exactly one local minimum in (0,1) | |
two distinct local minima in (0,1) |
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)
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
\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
0 | |
\infty | |
e^{sa} | |
e^{-sa} |
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
\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}
The integrating factor of the differential equation is
\left ( 1-x^{2} \right )^{-3/4} | |
\left ( 1-x^{2} \right )^{-1/4} | |
\left ( 1-x^{2} \right )^{-3/2} | |
\left ( 1-x^{2} \right )^{-1/2} |
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
(\sin x)\frac{dy}{dx}+y \cos x =1
subject to the condition y(\pi /2)=\pi /2, then y(\pi /6) is
0 | |
\frac{\pi}{6} | |
\frac{\pi}{3} | |
\frac{\pi}{2} |
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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