Question 1 |
The initial value problem
\frac{dy}{dt} +2y=0,\;\;y(0)=1
is solved numerically using the forward Euler's method with a constant and positive time step of \Delta t.
Let y_n represent the numerical solution obtained after n steps. The condition |y_{n+1}| \leq |y_n| is satisfied if and only if \Delta t does not exceed _____. (Answer in integer)
\frac{dy}{dt} +2y=0,\;\;y(0)=1
is solved numerically using the forward Euler's method with a constant and positive time step of \Delta t.
Let y_n represent the numerical solution obtained after n steps. The condition |y_{n+1}| \leq |y_n| is satisfied if and only if \Delta t does not exceed _____. (Answer in integer)
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
Solution of differential equation given by forward Euler's method is
y_{n+1}=y_{n}+h f\left(x_{n}, y_{n}\right)
Given: \frac{d y}{d t}+2 y=0
\Rightarrow \quad \frac{d y}{d t}=-2 y
and step size, \mathrm{h}=\Delta \mathrm{t}
\therefore \quad y_{n+1}=y_{n}+\Delta t\left(-2 y_{n}\right)
\mathrm{y}_{\mathrm{n}+1}=\mathrm{y}_{\mathrm{n}}(1-2 \Delta \mathrm{t})
\frac{y_{n+1}}{y_{n}}=1-2 \Delta t \;\;\;...(i)
Given:
\left|y_{n+1}\right| \leq\left|y_{n}\right|
\Rightarrow \quad\left|\frac{y_{n+1}}{y_{n}}\right| \leq 1
From equation (i)
\therefore \quad|1-2 \Delta t| \leq 1
Case (i) when (1-2 \Delta t) \gt 0 :
Then, \quad 1-2 \Delta t \leq 1
\Delta t \geq 0
Case (ii) when (1-2 \Delta t) \lt 0 :
then -(1-2 \Delta t) \leq 1
1-2 \Delta \mathrm{t} \geq-1
\Delta \mathrm{t} \leq 1
\therefore \quad Value of \Delta t: 0 \leq \Delta t \leq 1
y_{n+1}=y_{n}+h f\left(x_{n}, y_{n}\right)
Given: \frac{d y}{d t}+2 y=0
\Rightarrow \quad \frac{d y}{d t}=-2 y
and step size, \mathrm{h}=\Delta \mathrm{t}
\therefore \quad y_{n+1}=y_{n}+\Delta t\left(-2 y_{n}\right)
\mathrm{y}_{\mathrm{n}+1}=\mathrm{y}_{\mathrm{n}}(1-2 \Delta \mathrm{t})
\frac{y_{n+1}}{y_{n}}=1-2 \Delta t \;\;\;...(i)
Given:
\left|y_{n+1}\right| \leq\left|y_{n}\right|
\Rightarrow \quad\left|\frac{y_{n+1}}{y_{n}}\right| \leq 1
From equation (i)
\therefore \quad|1-2 \Delta t| \leq 1
Case (i) when (1-2 \Delta t) \gt 0 :
Then, \quad 1-2 \Delta t \leq 1
\Delta t \geq 0
Case (ii) when (1-2 \Delta t) \lt 0 :
then -(1-2 \Delta t) \leq 1
1-2 \Delta \mathrm{t} \geq-1
\Delta \mathrm{t} \leq 1
\therefore \quad Value of \Delta t: 0 \leq \Delta t \leq 1
Question 2 |
Consider the second-order linear ordinary differential equation
x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=0,\;\; x \geq 1
with the initial conditions
y(x=1)=6,\;\;\left.\begin{matrix} \frac{dy}{dx} \end{matrix}\right|_{x=1}=2
The value of y at x=2 equals _________. (Answer in integer)
x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=0,\;\; x \geq 1
with the initial conditions
y(x=1)=6,\;\;\left.\begin{matrix} \frac{dy}{dx} \end{matrix}\right|_{x=1}=2
The value of y at x=2 equals _________. (Answer in integer)
6 | |
9 | |
13 | |
17 |
Question 2 Explanation:
\frac{x d^{2} y}{d x^{2}}+\frac{x d y}{d x}-y=0, x \geq 1
This is an Euler-cauchy equation and has solutions of the form
\begin{aligned} & y=x^{n} \\ & y^{\prime}=n x^{n-1} \\ & y^{\prime \prime}=n(n-1) x^{n-2} \\ & x^{2}(n)(n-1) x^{n-2}+x\left(n x^{n-1}\right)-x^{n}=0 \\ & x^{n}[x(n-1)+n-1)=0 \\ & n^{2}-1=0 \\ & n=\pm 1 \\ & y_{1}=x \\ & y_{2}=\frac{1}{x} \\ & y=C_{1} x+\frac{C_{2}}{x} \end{aligned}
Given y(x=1)=6
\begin{aligned} 6 & =C_{1}(1)+\frac{C_{2}}{(1)} \\ C_{1}+C_{2} & =6 \;\;\;...(i)\\ \frac{d y}{d x} & =C_{1}-\frac{C_{2}}{x^{2}}=2 \\ x & =1 \end{aligned}
C_{1}=C_{2}=2
C_{1}=4, C_{2}=2
y(x=2)=4(2)+\frac{2}{(2)}=9
This is an Euler-cauchy equation and has solutions of the form
\begin{aligned} & y=x^{n} \\ & y^{\prime}=n x^{n-1} \\ & y^{\prime \prime}=n(n-1) x^{n-2} \\ & x^{2}(n)(n-1) x^{n-2}+x\left(n x^{n-1}\right)-x^{n}=0 \\ & x^{n}[x(n-1)+n-1)=0 \\ & n^{2}-1=0 \\ & n=\pm 1 \\ & y_{1}=x \\ & y_{2}=\frac{1}{x} \\ & y=C_{1} x+\frac{C_{2}}{x} \end{aligned}
Given y(x=1)=6
\begin{aligned} 6 & =C_{1}(1)+\frac{C_{2}}{(1)} \\ C_{1}+C_{2} & =6 \;\;\;...(i)\\ \frac{d y}{d x} & =C_{1}-\frac{C_{2}}{x^{2}}=2 \\ x & =1 \end{aligned}
C_{1}=C_{2}=2
C_{1}=4, C_{2}=2
y(x=2)=4(2)+\frac{2}{(2)}=9
Question 3 |
Which one of the options given is the inverse Laplace transform of \frac{1}{s^3-s}
u(t)denotes the unit-step function.
u(t)denotes the unit-step function.
\left ( -1+\frac{1}{2}e^{-t}+\frac{1}{2}e^{t} \right )u(t) | |
\left (\frac{1}{3}e^{-t}-e^{t} \right )u(t) | |
\left ( -1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)} \right )u(t-1) | |
\left ( -1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)} \right )u(t-1) |
Question 3 Explanation:
Inverse Laplace of \frac{1}{s^{3}-s}
\qquad L^{-1}\left[\frac{1}{s\left(s^{2}-1\right)}\right](u(t))
\begin{aligned} & =\mathrm{L}^{-1}\left[\frac{1}{s} \times \frac{1}{(s-1)} \times \frac{1}{(s+1)}\right] u(t) \\ & =L^{-1}\left[\frac{-1}{s}+\frac{1}{2} \times \frac{1}{(s-1)}+\frac{1}{2(s+2)}\right] u(t) \\ & =\left[\frac{e^{t}}{2}+\frac{e^{-t}}{2}-1\right] u(t) \end{aligned}
\qquad L^{-1}\left[\frac{1}{s\left(s^{2}-1\right)}\right](u(t))
\begin{aligned} & =\mathrm{L}^{-1}\left[\frac{1}{s} \times \frac{1}{(s-1)} \times \frac{1}{(s+1)}\right] u(t) \\ & =L^{-1}\left[\frac{-1}{s}+\frac{1}{2} \times \frac{1}{(s-1)}+\frac{1}{2(s+2)}\right] u(t) \\ & =\left[\frac{e^{t}}{2}+\frac{e^{-t}}{2}-1\right] u(t) \end{aligned}
Question 4 |
For the exact differential equation, \frac{du}{dx}=\frac{-xu^2}{2+x^2u} , which one of the following is the solution ?
u^2+2x^2=\text{Constant} | |
xu^2+u=\text{Constant} | |
\frac{1}{2}x^2 u^2+2u=\text{Constant} | |
\frac{1}{2} u x^2+2x=\text{Constant} |
Question 4 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
\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 5 |
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?
\phi (s)=0 at all the three values of s satisfying s^3+\sigma ^3=0 | |
\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^2 \phi (s)}{ds^2}=0 \text{ at }s=-\sigma | |
\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^4 \phi (s)}{ds^4}=0 \text{ at }s=-\sigma | |
\phi (s)=0, \text{ and }\frac{d^3 \phi (s)}{ds^3}=0 \text{ at }s=-\sigma |
Question 5 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.
Therefore, \varphi (s)=(s+\sigma )^3\psi (s)
It satisfies all the conditions in option (B) is correct.
There are 5 questions to complete.