# Differential Equations

 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)
 A 0 B 1 C 2 D 3
GATE ME 2023   Engineering Mathematics
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 $\Delta t: 0 \leq \Delta t \leq 1$
 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)
 A 6 B 9 C 13 D 17
GATE ME 2023   Engineering Mathematics
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}
$C_{1}=C_{2}=2$
$C_{1}=4, C_{2}=2$
$y(x=2)=4(2)+\frac{2}{(2)}=9$

 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.
 A $\left ( -1+\frac{1}{2}e^{-t}+\frac{1}{2}e^{t} \right )u(t)$ B $\left (\frac{1}{3}e^{-t}-e^{t} \right )u(t)$ C $\left ( -1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)} \right )u(t-1)$ D $\left ( -1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)} \right )u(t-1)$
GATE ME 2023   Engineering Mathematics
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}
 Question 4
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 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
$\int_{\text{u constant}}Mdx+\int_{\text{term without 'x'}}Ndx=c$
$\Rightarrow \int (xu^2)dx+\int 2du=c$
$\therefore \; \frac{x^2u^2}{2}+2u=constant$
 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?
 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 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.

There are 5 questions to complete.

### 1 thought on “Differential Equations”

1. There is a mistake in the solution Q45, the correct answer is D(0.99).