# Ordinary Differential Equation

 Question 1
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 1 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 2
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 2 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 3
An ordinary differential equation is given below

$6\frac{d^2y}{dx^2}+\frac{dy}{dx}-y=0$

The general solution of the above equation (with constant $C_1 \; and \; C_2$), is
 A $y(x)=C_1 e^{-\frac{x}{3}}+C_2 e^{\frac{x}{2}}$ B $y(x)=C_1 e^{\frac{x}{3}}+C_2 e^{-\frac{x}{2}}$ C $y(x)=C_1 xe^{-\frac{x}{3}}+C_2 e^{\frac{x}{2}}$ D $y(x)=C_1 e^{-\frac{x}{3}}+C_2 xe^{\frac{x}{2}}$
GATE CE 2020 SET-2   Engineering Mathematics
Question 3 Explanation:
\begin{aligned} \frac{6d^2y}{dx^2}+\frac{dy}{dx}-y &=0 \\ (6D^2+D-1)y&=0 \\ 6D^2+3D-2D-1 &=0 \\ (2D+1)(3D-1)&=0 \\ D&=\frac{-1}{2}, D=\frac{1}{3} \\ y&=C_1e^{x/3}+C_2e^{-x/2} \end{aligned}
 Question 4
The following partial differential equation is defined for $u:u(x,y)$

$\frac{\partial u}{\partial y}=\frac{\partial^2 u}{\partial x^2}; \; \; y\geq 0;\;x_1\leq x\leq x_2$

The set of auxiliary conditions necessary to solve the equation uniquely, is
 A three initial conditions B three boundary conditions C two initial conditions and one boundary condition D one initial condition and two boundary conditions
GATE CE 2020 SET-2   Engineering Mathematics
Question 4 Explanation:
Given: DE is $\frac{\partial u}{\partial y}=\frac{\partial^2 u}{\partial x^2}; \; y\geq 0;\; x_1 \leq x\leq x_2$
$\because$ y is given as $\geq 0$ so we take it as time. Hence, above equation is nothing but one-D heat equation which requires one initial condition and two boundary condition.
 Question 5
The ordinary differential equation $\frac{d^2u}{dx^2}-2x^2u+\sin x=0$ is
 A linear and homogeneous B linear and nonhomogeneous C nonlinear and homogeneous D nonlinear and nonhomogeneous
GATE CE 2020 SET-2   Engineering Mathematics
Question 5 Explanation:
Its solution is of the type u=f(x), i.e., dependent variable is u.
Hence, given equation is Linear and Non-Homogeneous.
 Question 6
A continuous function $f(x)$ is defined. If the third derivative at $x_i$ is to be computed by using the fourth order central finite-divided-difference scheme (the step length = h), the correct formula is
 A A B B C C D D
GATE CE 2020 SET-1   Engineering Mathematics
Question 6 Explanation:
$\left. \begin{matrix}\frac{\partial^3 u }{\partial x^3} \end{matrix}\right|_{x_i}=\frac{-u_{i+3} +8u_{i+2}-13u_{i+1}+13u_{i-1}-8u_{i-2}+u_{i-3}}{8\Delta h^3}$
$f''(x_i)=\frac{-f(x_{i+3})+8f(x_{i+2}-13f(x_{i+1}+13f(x_{i-1}-8f(x_{i-2}+f(x_{i-3}}{8h^3}$
 Question 7
For the Ordinary Differential Equation $\frac{d^2x}{dt^2}-5\frac{dx}{dt}+6x=0$, with initial condition $x(0)=0$ and $\frac{dx}{dt}(0)=10$, the solution is
 A $-5e^{2t}+6e^{3t}$ B $5e^{2t}+6e^{3t}$ C $-10e^{2t}+10e^{3t}$ D $10e^{2t}+10e^{3t}$
GATE CE 2020 SET-1   Engineering Mathematics
Question 7 Explanation:
A.E. is $m^2-5m+6=0\Rightarrow m=2,3$
So $C_f=C_1e^{2t}+C_2e^{3t}$
PI =0 and G. Solution is $x=CF+PI=C_1e^{2t}+C_2e^{3t}$
and $\frac{dx}{dt}=2C_1e^{2t}+3C_2e^{3t}$
Now, using initial condition we get $C_1=-10, C_2=10$
$x=-10e^{2t}+10e^{3t}$
 Question 8
The area of an ellipse represented by an equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
 A $\frac{\pi a b}{4}$ B $\frac{\pi a b}{2}$ C $\pi a b$ D $\frac{4 \pi a b}{3}$
GATE CE 2020 SET-1   Engineering Mathematics
Question 8 Explanation:

\begin{aligned} \text{Area} &=\int \int (1)dydx \\ &=\int_{x=-a}^{a}\int_{y=-\frac{b}{a}}^{+\frac{b}{a}}(1)dydx \\ &=4\int_{x=0}^{a} \int_{y=0}^{\frac{b}{a}\sqrt{a^2-x^2}}(1)dydx\\ &= 4 \int_{x=0}^{a}\int_{y=0}^{\frac{b}{a}\sqrt{a^2-x^2}} dx\\ &= \pi ab \end{aligned}
 Question 9
In the following partial differential equation, $\theta$ is a function of t and z, and D and K are functions of $\theta$

$D(\theta )\frac{\partial^2 \theta }{\partial z^2}+\frac{\partial K(\theta )}{\partial z}-\frac{\partial \theta }{\partial t}=0$

The above equation is
 A a second order linear equation B a second degree linear equation C a second order non-linear equation D a second degree non-linear equation
GATE CE 2020 SET-1   Engineering Mathematics
Question 9 Explanation:
$\because \;\;1^{st}$ term of given D. Equation contains product of dependent variable with it's derivative, so it is non-linear and also we have 2nd order derivative so it's order is two
i.e., 2nd order non linear equation.
 Question 10
An ordinary differential equation is given below.
$\left ( \frac{dy}{dx} \right )(x lnx)=y$
The solution for the above equation is
(Note: K denotes a constant in the options)
 A $y=Kx \ln x$ B $y=Kxe^x$ C $y=Kxe^{-x}$ D $y=K \ln x$
GATE CE 2019 SET-2   Engineering Mathematics
Question 10 Explanation:
\begin{aligned} \frac{dy}{dx}(x \ln x) &=y \\ \frac{dy}{y}&=\frac{dx}{x \ln x} \\ \int \frac{dy}{y}&=\int \frac{1}{x \ln x}dx+ \ln K \\ \ln x&=t \\ \frac{1}{x}dx&=dt \\ \ln y&= \int \frac{dy}{t}+ \ln K\\ \ln y&=\ln t+ \ln K \\ \ln y&=Kt \\ y&=K \ln x \end{aligned}
There are 10 questions to complete.