# Ordinary Differential Equation

 Question 1
The solution of the differential equation.

$\frac{d^{3} y}{d x^{3}}-5.5 \frac{d^{2} y}{d x^{2}}+9.5 \frac{d y}{d x}-5 y=0$

is expressed as $y=C_{1} e^{2.5 x}+C_{2} e^{\alpha . x}+C_{3} e^{\beta x}$, where $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ and $\alpha$ and $\beta$ are constants, with $\alpha$ and $\beta$ being distinct and not equal to 2.5. which of the following options is correct for the values of $\alpha$ and $\beta$ ?
 A 1 and 2 B -1 and -2 C 2 and 3 D -2 and -3
GATE CE 2023 SET-2   Engineering Mathematics
Question 1 Explanation:
The differential eqn. can be written as :
$D^{3}-5.5 D^{2}+9.5 D-5=0$
Solving we get,
$D=1,2,2.5$
Hence, the solution is given as
$y=C_{1} e^{2.5 x}+C_{2} e^{1 x}+C_{3} e^{2 x}$
Comparing we get,
\begin{aligned} & \alpha=1 \\ & \beta=2 \end{aligned}
Hence, option (A) is correct.
 Question 2
The differential equation,

$\frac{\mathrm{du}}{\mathrm{dt}}+2 \mathrm{tu}^{2}=1$

is solved by employing a backward differnce scheme within the finite difference framework. The value of $u$ at the $(n-1)^{\text {th }}$ time-step, for some $n$, is 1.75. The corresponding time $(\mathrm{t})$ is $3.14 \mathrm{~s}$. Each time step is $0.01 \mathrm{~s}$ long. Then, the value of $(u_{n}-u_{n-1})$ is ____ (round off to three decimal places).
 A -0.125 B 0.125 C -0.182 D 0.182
GATE CE 2023 SET-1   Engineering Mathematics
Question 2 Explanation:
\begin{aligned} \frac{d u}{d t} & =1-2 t u^{2} \\ \frac{d u}{d t} & =f(t, u) \\ \Rightarrow \quad f(t, u) & =1-2 \mathrm{tu}^{2} \end{aligned}
Eulers backward difference scheme
\begin{aligned} u_{n}-u_{n-1} & =h f\left(t_{n}, u_{n}\right)=h\left(1-2 t_{n} u n^{2}\right) \\ & =0.01\left(1-2 \times 3.14 \times 1.75^{2}\right) \\ & =-0.1822 \end{aligned}

 Question 3
In the differential equation $\frac{d y}{d x}+a x y=0$, $a$ is positive constant. If $y=1.0$ at $x=0.0$, and $y=0.8$ at $x=1.0$, the value of $a$ is ___ (rounded off to three decimal places).
 A 0.446 B 0.124 C 0.745 D 0.852
GATE CE 2023 SET-1   Engineering Mathematics
Question 3 Explanation:
\begin{aligned} \frac{d y}{d x}+\alpha x y & =0 \\ \frac{d y}{d x} & =-\alpha x y \\ \frac{d y}{y} & =-\alpha x d x \\ \frac{d y}{y} & =-\alpha \int x d x+C \\ \text { In } y & =-\frac{\alpha x^{2}}{2}+C \\ \text { Put } y&=1 \text { at } =0 \\ \ln 1 & =0+C \\ C & =0 \\ \ln y & =-\frac{\alpha x^{2}}{2} \end{aligned}
Put $y=0.8$ at $x=1$
\begin{aligned} \ln (0.8) & =-\frac{\alpha}{2} \\ \alpha & =-2 \ln (0.8) \\ \alpha & =0.4462 \end{aligned}
 Question 4
Consider the differential equation
$\frac{dy}{dx}=4(x+2)-y$
For the initial condition $y = 3$ at $x = 1$, the value of $y$ at $x = 1.4$ obtained using Euler's method with a step-size of 0.2 is _________. (round off to one decimal place)
 A 5.4 B 6.4 C 2.8 D 4.2
GATE CE 2022 SET-1   Engineering Mathematics
Question 4 Explanation:
\begin{aligned} x_0&=1\rightarrow y_0=3\\ x_1&=1.2\rightarrow y_1=?\\ x_2&=1.4\rightarrow y_2=?\\ \end{aligned}
Using Euler's forward methods
\begin{aligned} y_n&=y_n+hf(x_n,y_n)\\ y_1&=y_0+hf(x_0,y_0)\\ &=3+0.2 \times (4(1+2)-3)\\ &=4.8\\ y_2&=y_1+hf(x_1,y_1)\\ &=4.8+0.2 \times (4(1.2+2)-4.8)\\ &=6.4\\ \end{aligned}
 Question 5
For the equation
$\frac{d^3y}{dx^3}+x\left ( \frac{dy}{dx} \right )^{\frac{3}{2}}+x^2y=0$
the correct description is
 A an ordinary differential equation of order 3 and degree 2. B an ordinary differential equation of order 3 and degree 3. C an ordinary differential equation of order 2 and degree 3. D an ordinary differential equation of order 3 and degree 3/2.
GATE CE 2022 SET-1   Engineering Mathematics
Question 5 Explanation:
$\frac{d^3y}{dx^3}+x\left ( \frac{dy}{dx} \right )^{3/2}+x^2y=0$
Power of $\left ( \frac{dy}{dx} \right )$ is fractional, make it integer.
$\frac{d^3y}{dx^3}+x^2y=-x\left ( \frac{dy}{dx} \right )^{3/2}$
$\left (\frac{d^3y}{dx^3}+x^2y \right )^2=x^2\left ( \frac{dy}{dx} \right )^{3}$
Now order = 3 and degree = 2

There are 5 questions to complete.