Differential Equations

Question 1
Consider the initial value problem below. The value of y at x = ln \;2. (rounded off to 3 decimal places) is ________ .

\frac{dy}{dx}=2x-y,\; y(0)=1
A
1.386
B
0.886
C
0.452
D
0.642
GATE EE 2020   Engineering Mathematics
Question 1 Explanation: 
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{d} x}&=2x-y \\ y(0)&=1, \; y \text{ at } x=\ln 2\\ \frac{\mathrm{d} y}{\mathrm{d} x}+y&=2x \\ P&=1, \;\; Q=2x \\ I.F.&=e^{\int Pdx}=e^{\int 1dx}=e^{x}\\ \text{Solution, } Y(I.F)&=\int Q(I.F.)dx\\ye^{x}&=\int 2x\cdot e^{x}dx \\&=2(xe^{x}-e^{x})+C \\ y&=2x-2+ce^{-x} \\ y(0)&=1 \\ 1&=0-2+C \\ C&=3 \\ \therefore \; \; y&=2x-2+3e^{-x} \\ \text{At }x&= \ln 2 \\ y&=2(\ln 2)-2+3e^{-\ln 2} \\ &=1.386-2+\frac{3}{2}\\ &=0.886\end{aligned}
Question 2
The partial differential equation
\frac{\partial^2 u}{\partial t^2}-c^2\left ( \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} \right )=0; where c\neq 0
is known as
A
heat equation
B
wave equation
C
Poisson's equation
D
Laplace equation
GATE EE 2019   Engineering Mathematics
Question 3
Consider a system governed by the following equations
\frac{dx_{1}(t)}{dt}=x_{2}(t)-x_{1}(t)
\frac{dx_{2}(t)}{dt}=x_{1}(t)-x_{2}(t)
The initial conditions are such that x_{1} \lt x_{2}(0) \lt \infty. Let x_{1f}=\lim_{t\rightarrow \infty }x_{1}(t) and x_{2f}=\lim_{t\rightarrow \infty }x_{2}(t). Which one of the following is true?
A
x_{1f} \lt x_{2f} \lt \infty
B
x_{2f} \lt x_{1f} \lt \infty
C
x_{1f}=x_{2f} \lt \infty
D
x_{1f}=x_{2f}=\infty
GATE EE 2018   Engineering Mathematics
Question 3 Explanation: 
\begin{aligned} (D+1)x_1=x_2\\ \Rightarrow \; (D+1)x_1-x_2=0\\ (D+1)x_2=x_1\\ \Rightarrow \;-x_1+(D+1)x_2=0\\ x_{1f}=\lim_{t \to \infty }x_1(t)\\ x_{2f}=\lim_{t \to \infty }x_2(t)\\ (D+1)x_1-x_2=0\\ -(D+1)x_1+(D+1)^2x_2=0\\ ((D+1)^2-1)x_2=0\\ (D^2+2D)x_2=0\\ D^2+2D=0\\ D(D+2)=0\\ \therefore \;\; x^2=C_1+C_2e^{-2t}\;\;...(1)\\ D=0,-2\\ x_{2f}=\lim_{t \to \infty }C_1+C_2e^{-2t}\\ =C_1\\ (D+1)x_1-(D+1)x_2=0\\ -x_1+(D+1)x_2=0\\ ((D+1)^2-1)x_1=0\\ (D^2+2D)x_1=0\\ x_{1f}=\lim_{t \to \infty }C_1+C_2e^{-2t}\\ =C_1\\ x_{1f}=C_1 \end{aligned}
Question 4
Consider the differential equation (t^{2}-81)\frac{dy}{dt}+5ty=sin(t) with y(1)=2\pi. There exists a unique solution for this differential equation when t belongs to the interval
A
(-2,2)
B
(-10,10)
C
(-10,2)
D
(0,10)
GATE EE 2017-SET-1   Engineering Mathematics
Question 4 Explanation: 
The differentail equation,
\begin{aligned} (t^2-81)\frac{dy}{dt}+5+y&= \sin t\\ \frac{dy}{dt}+\frac{5t}{t^2-81}y&=\frac{\sin t }{t^2-81}\\ P&=\frac{5t}{t^2-81}\\ Q&=\frac{\sin t}{t^2-81}\\ I.F&=e^{\int \rho dt}=e^{\int \frac{5t}{t^2-81}dt}\\ &=e^{\frac{5}{2}\int \frac{2t}{t^2-81}dt}=e^{\frac{5}{2}\ln ({t^2-81})}\\ &=e^{\ln {(t^2-81)^{5/2}}}=(t^2-81)^{5/2}\\ \text{Solution is }&\\ y(t^2-81)^{5/2}&=\int \frac{\sin t}{t^2-81}(t^2-81)^{5/2}dt\\ y&=\frac{\int (\sin t)(t^2-81)^{3/2}dt}{(t^2-81)^{5/2}}+\frac{C}{(t^2-81)^{5/2}}\\ &\text{The solution exist for } t\neq -9, t\neq +9 \end{aligned}
Only option (A) not cover 9 and -9. Hence it is correct.
Question 5
Let y(x) be the solution of the differential equation \frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+4y=0 with initial conditions y(0)=0 and \frac{dy}{dx}|_{x=0}=1. Then the value of y(1) is _________.
A
2.50
B
5.65
C
7.38
D
9.36
GATE EE 2016-SET-2   Engineering Mathematics
Question 5 Explanation: 
\begin{aligned} \text{A.E. }&m^2-4m+4=0\\ m&=2,2\\ y&=(C_1+C_2x)e^{2x}\\ y(0)&=0\Rightarrow C_1=0\\ y&=C_2xe^{2x}\\ y'&=C_2e^{2x}+2C_2xe^{2x}\\ y'(0)&=1\\ C_2&=1\\ y&=xe^{2x}\\ y(1)&=e^2\\ &=7.38 \end{aligned}
Question 6
The solution of the differential equation, for t \gt 0, y''(t)+2y'(t)+y(t)=0 with initial conditions y(0) = 0 and y'(0) = 1, is (u(t) denotes the unit step function),
A
te^{-t}u(t)
B
(e^{-t}-te^{-t})u(t)
C
(-e^{-t}+te^{-t})u(t)
D
e^{-t}u(t)
GATE EE 2016-SET-2   Engineering Mathematics
Question 6 Explanation: 
The differentail equation is
y'(t)+2y'(t)+y(t)=0
So, (s^2y(s)-sy(0)-y'(0))+2[sy(s)-y(0)]+y(s) =0
So, y(s)=\frac{sy(0)+y'(0)+2y(0)}{s^2+2s+1}
Given that y'(0)=1, y(0)=0
So, y(s)=\frac{1}{(s+1)^2}
So, y(t)=te^{-t}u(t)
Question 7
A solution of the ordinary differential equation \frac{d^{2}y}{dt^{2}}+5\frac{dy}{dt}+6y=0 is such that y(0)=2 and y(1)= -\frac{1-3e}{e^{3}}. The value of \frac{dy}{dt}(0) is _______
A
-1
B
-8
C
3
D
-3
GATE EE 2015-SET-1   Engineering Mathematics
Question 7 Explanation: 
\begin{aligned} D^2+5D+6&=0 \\ D &=-1,-3 \\ y(t)&=c_1e^{-2t}+c_2e^{-3t} \\ \text{Given, } y(0)&=2 \\ \Rightarrow \; c_1+c_2 &=2\;\;...(i) \\ y(1)&=-\left ( \frac{1-3e}{e^3} \right ) \\ \Rightarrow \;\; \frac{c_1}{e^2}+\frac{c_2}{e^3} &= -\left ( \frac{1-3e}{e^3} \right ) \\ \Rightarrow \;\;ec_1+c_2 &=3e-1\;\;\;...(ii) \\ \text{Now solving equation} &\text{ (i) and (ii), we get} \\ c_1=3, &c_2=-1\\ \text{Substituting in }y(t),& we get \\ y(t)&=3e^{-2t}-e^{-3t}\\ \text{Now, }\frac{dy}{dt}&=-6e^{-2t}+3e^{-3t}\\ \left ( \frac{dy}{dt} \right )_{t=0}&=-6+3=-3 \end{aligned}
Question 8
Consider the differential equation x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-y=0. Which of the following is a solution to this differential equation for x \gt 0 ?
A
e^{x}
B
x^{2}
C
1/x
D
ln x
GATE EE 2014-SET-2   Engineering Mathematics
Question 8 Explanation: 
\begin{aligned} x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y&=0\\ \text{Let, }x=e^z\leftrightarrow z&=\log x\\ s\frac{d}{dx}=xD=\theta &=\frac{d}{dz}\\ x^2D^2&=\theta (\theta -1)\\ (x^2D^2+xD-1)y&=0\\ [\theta (\theta -1)+\theta -1]y&=0\\ (\theta ^2-\theta +\theta -1)&=0\\ (\theta ^2-1)y&=0\\ \text{Auxiliary equation is }m^2-1&=0\\ m&=\pm 1\\ \text{CF is }C_1e^{-z}+C_2e^z&\\ \text{Solution is }y&=C_1e^{-z}+C_2e^z\\ y&=C_1x^{-1}+C_2x\\ y&=C_1\frac{1}{x}+C_2x \end{aligned}
One independent solution is \frac{1}{x}
Another independent solution is x.
Question 9
The solution for the differential equation
\frac{d^{2}x}{dt^{2}}=-9x
with initial conditions x(0)=1 and \frac{dx}{dt}|_{t=0}=1, is
A
t^{2}+t+1
B
sin 3t+ \frac{1}{3} cos3t+\frac{2}{3}
C
\frac{1}{3} sin 3t+cos3t
D
cos3t+t
GATE EE 2014-SET-1   Engineering Mathematics
Question 9 Explanation: 
\begin{aligned} \frac{d^2x}{dt^2}&=-9x\\ \frac{d^2x}{dt^2}+9x&=0\\ (D^2+9)x &=0 \end{aligned}
Auxiliary equation is m^2+9=0
\begin{aligned} m&= \pm 3i\\ x&= C_1 \cos 3t+C_2 \sin 3t\;\;...(i)\\ x(0) &=1\;\;i.e.\;x\rightarrow 1\; when t\rightarrow 0 \\ 1&=C_1 \\ \frac{dx}{dt}&=-3C_1 \sin 3t+3C_2 \cos 3t\;\;...(ii) \\ x'(0)&=1\;\; i.e.\;x'\rightarrow 1\; when \; t\rightarrow 0 \\ 1&=3C_2 \\ C_2&=\frac{1}{3}\\ \therefore \;x&=\cos 3t+\frac{1}{3}\sin 3t \end{aligned}
Question 10
With initial condition x(1)=0.5, the solution of the differential equation t\frac{dx}{dt}+x=t, is
A
x=t-\frac{1}{2}
B
x=t^{2}-\frac{1}{2}
C
x=\frac{t^{2}}{2}
D
x=\frac{t}{2}
GATE EE 2012   Engineering Mathematics
Question 10 Explanation: 
The given differential equation is \frac{tdx}{dt}+x=t with initial condition x(t)=\frac{1}{2} which is same as \frac{dx}{dt}+\frac{x}{t}=1
Which is a linear differential equation
\frac{dx}{dt}+P{x}=Q
Where, P=\frac{1}{t} and Q=1
Integrating factor =e^{\int P\;dt}=e^{\int \frac{1}{t}dt}=e^{\log _{e}t}=t
Solution is
\begin{aligned} x\cdot (IF) &=\int Q\cdot (IF)dt +C\\ x\cdot t&= \int 1\cdot t\cdot dt+C\\ xt&= \frac{t^2}{2}+C\\ x&= \frac{t}{2}+\frac{C}{t}\\ \text{Put } x(1)&=\frac{1}{2} \\ \Rightarrow \; \frac{1}{2}+\frac{C}{1}&=\frac{1}{2} \\ \Rightarrow \; C&=0 \\ \text{So, }x&=\frac{t}{2}\text{ is the solution.} \end{aligned}
There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.