Question 1 |
Consider the following partial differential equation (PDE)
a\frac{\partial^2 f(x,y)}{\partial x^2} +b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)
, where aand b are distinct positive real numbers. Select the combination(s) of values of the real parameters \xi and \eta such that f(x,y)=e^{(\xi x+\eta y)} is a solution of the given PDE.
a\frac{\partial^2 f(x,y)}{\partial x^2} +b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)
, where aand b are distinct positive real numbers. Select the combination(s) of values of the real parameters \xi and \eta such that f(x,y)=e^{(\xi x+\eta y)} is a solution of the given PDE.
\xi =\frac{1}{\sqrt{2a}},\eta =\frac{1}{\sqrt{2b}} | |
\xi =\frac{1}{\sqrt{a}},\eta =0 | |
\xi =0,\eta =0 | |
\xi =\frac{1}{\sqrt{a}},\eta =\frac{1}{\sqrt{b}} |
Question 1 Explanation:
Given
\begin{aligned} f(x,y)&=e^{(\xi x+\eta y)}\\ \frac{\partial f(x,y)}{\partial x}&=\xi \cdot e^{(\xi x+\eta y)}\\ \frac{\partial^2 f(x,y)}{\partial x^2}&=\xi ^2\cdot e^{(\xi x+\eta y)}\\ \end{aligned}
Similarly,
\frac{\partial^2 f(x,y)}{\partial y^2}=\eta ^2\cdot e^{(\xi x+\eta y)}
Now, as given
a \frac{\partial^2 f(x,y)}{\partial x^2}+b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)
or
a \times \xi ^2\cdot e^{(\xi x+\eta y)}+b \times \eta ^2\cdot e^{(\xi x+\eta y)}= e^{(\xi x+\eta y)}
a\cdot \xi ^2+b\cdot \eta ^2=1
Thus, \xi =\frac{1}{\sqrt{2a}}, \eta = \frac{1}{\sqrt{2b}}
(\xi =\frac{1}{\sqrt{a}}, \eta = 0) and (\xi =0, \eta = \frac{1}{\sqrt{b}}) satisfy the above result.
\begin{aligned} f(x,y)&=e^{(\xi x+\eta y)}\\ \frac{\partial f(x,y)}{\partial x}&=\xi \cdot e^{(\xi x+\eta y)}\\ \frac{\partial^2 f(x,y)}{\partial x^2}&=\xi ^2\cdot e^{(\xi x+\eta y)}\\ \end{aligned}
Similarly,
\frac{\partial^2 f(x,y)}{\partial y^2}=\eta ^2\cdot e^{(\xi x+\eta y)}
Now, as given
a \frac{\partial^2 f(x,y)}{\partial x^2}+b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)
or
a \times \xi ^2\cdot e^{(\xi x+\eta y)}+b \times \eta ^2\cdot e^{(\xi x+\eta y)}= e^{(\xi x+\eta y)}
a\cdot \xi ^2+b\cdot \eta ^2=1
Thus, \xi =\frac{1}{\sqrt{2a}}, \eta = \frac{1}{\sqrt{2b}}
(\xi =\frac{1}{\sqrt{a}}, \eta = 0) and (\xi =0, \eta = \frac{1}{\sqrt{b}}) satisfy the above result.
Question 2 |
Consider the differential equation given below.
\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}
The integrating factor of the differential equation is
\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}
The integrating factor of the differential equation is
\left ( 1-x^{2} \right )^{-3/4} | |
\left ( 1-x^{2} \right )^{-1/4} | |
\left ( 1-x^{2} \right )^{-3/2} | |
\left ( 1-x^{2} \right )^{-1/2} |
Question 2 Explanation:
\begin{aligned} \frac{d y}{d x}+\frac{x}{1-x^{2}} y&=x \sqrt{y}, \quad \text { IF }=?\\ \text{Divided by }\sqrt{y}\\ \frac{1}{\sqrt{y}} \frac{d y}{d x}+\frac{x}{1-x^{2}} \sqrt{y}&=x \\ 2 \frac{d u}{d x}+\frac{x}{1-x^{2}} u&=x\\ \text{Let }\qquad x \sqrt{y}&=u\\ \frac{1}{2 \sqrt{v}} \frac{d y}{d x}&=\frac{d u}{d x}\\ \Rightarrow \qquad \frac{d u}{d x}+\frac{x}{2\left(1-x^{2}\right)} u&=\frac{x}{2} \rightarrow \text{ lines diff. equ.} \\ \text { I. } F&=e^{\int \frac{x}{2\left(1-x^{2}\right)} d x}=e^{-\frac{1}{4} \log \left(1-x^{2}\right)}&=e^{\log \left(1-x^{2}\right) \frac{-1}{4}} \\ \text { I.F }&=\frac{1}{\left(1-x^{2}\right)^{\frac{1}{4}}} \end{aligned}
Question 3 |
Which one of the following options contains two solutions of the differential equation \frac{dy}{dx}=(y-1)x?
ln |y-1|=0.5x^2 +C and y=1 | |
ln |y-1|=2x^2 +C and y=1 | |
ln |y-1|=0.5x^2 +C and y=-1 | |
ln |y-1|=2x^2 +C and y=-1 |
Question 3 Explanation:
\begin{aligned} \frac{dy}{dx}&=(y-1)x\\ \frac{dy}{y-1}&=x dx\\ \text{Integrating}&\text{ both side, we get}\\ \int \frac{1}{y-1}dy&=\int xdx\\ \ln |y-1|&=\frac{x^2}{2}+C\; \; \;...(i)\\ \text{Now, }&\\ \frac{dy}{dx}&=0 \text{ when y=1}\\ \text{It means }y&= \text{constant C}\\ \therefore \; y&=1\;\;\;...(ii) \end{aligned}
So, equation (i) and (ii) are two different solution of given differential equation.
So, equation (i) and (ii) are two different solution of given differential equation.
Question 4 |
The general solution of \frac{d^2y}{dx^2}-6\frac{dy}{dx}+9y=0 is
y=C_1e^{3x}+C_2e^{-3x} | |
y=(C_1+C_2x)e^{-3x} | |
y=(C_1+C_2x)e^{3x} | |
y=C_1e^{3x} |
Question 4 Explanation:
Taking \frac{\mathrm{d} }{\mathrm{d} x}=D
Given, D^{2}-6D+9=0
(D-3)^2=0
D=3,3
So, Solution of the given Differential equation
y=(C_{1}+C_{2}x)e^{3x}
Given, D^{2}-6D+9=0
(D-3)^2=0
D=3,3
So, Solution of the given Differential equation
y=(C_{1}+C_{2}x)e^{3x}
Question 5 |
Consider the homogeneous ordinary differential equation
x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+3y=0, \; x \gt 0
with y(x) as a generalsolution. Given that
y(1)=1 and y(2)=14
the value of y(1.5), rounded off to two decimal places, is ______.
x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+3y=0, \; x \gt 0
with y(x) as a generalsolution. Given that
y(1)=1 and y(2)=14
the value of y(1.5), rounded off to two decimal places, is ______.
1.25 | |
4.25 | |
3.75 | |
5.25 |
Question 5 Explanation:
\begin{aligned} \left(x^{2} D^{2}-3 x D+3\right)&=0 \\ (0(0-1)-30+3) y&=0 \\ \left(0^{2}-40+3\right) y&=0 \\ \text { AE is }\quad m^{2}-4 m+3&=0 \\ m&=1.3 \\ CF&=C_{1}x+C_{2}x^{3} \\ \text { Solution is } \quad y&=C_{1}x+C_{2}x^{3} \\ y(1)=1\quad 1&=C_{1}+C_{2} \\ y(2)=14\quad 14&=2 C_{1}+8 C_{2} \\ \text{From (il) and (ii). }C_{1}&=-1, \quad C_{2}=2 \\ \therefore\quad y&=-x+2 x^{3} \\ y(1.5)&=-1.5+2(1.5)^{3}=5.25 \end{aligned}
Question 6 |
The families of curves represented by the solution of the equation
\frac{dy}{dx}=-\left (\frac{x}{y} \right )^n
for n = -1 and n = +1, respectively, are
\frac{dy}{dx}=-\left (\frac{x}{y} \right )^n
for n = -1 and n = +1, respectively, are
Parabolas and Circles | |
Circles and Hyperbolas | |
Hyperbolas and Circles | |
Hyperbolas and Parabolas |
Question 6 Explanation:
\begin{aligned} \frac{d y}{d x} &=-\left(\frac{x}{y}\right)^{n} \\ n=-1\quad\quad \frac{d y}{d x} &=-\frac{y^{\prime}}{x} \\ \frac{d y}{y} &=-\frac{d x}{x} \\ \int \frac{1}{y} d y &=-\int \frac{1}{x} d x \\ \ln y &=-\ln x+\ln c \\ \ln (y x) &=\ln c \end{aligned}
x y=c \quad (Represents rectangular hyporbola)
\begin{aligned} n=1, \quad \frac{d y}{d x}&=-\frac{x}{y} \\ y d y &=-x d x \\ y d y &=-\int x d x \\ \frac{y^{2}}{2} &=-\frac{x^{2}}{2}+c \end{aligned}
x^{2}+y^{2}=2 c \quad (Represents family of circles)
x y=c \quad (Represents rectangular hyporbola)
\begin{aligned} n=1, \quad \frac{d y}{d x}&=-\frac{x}{y} \\ y d y &=-x d x \\ y d y &=-\int x d x \\ \frac{y^{2}}{2} &=-\frac{x^{2}}{2}+c \end{aligned}
x^{2}+y^{2}=2 c \quad (Represents family of circles)
Question 7 |
The position of a particle y(t) is described by the differential equation:
\frac{d^{2}y}{dt^{2}}=-\frac{dy}{dt}-\frac{5y}{4}
The initial conditions are y(0) = 1 and \left.\begin{matrix} \frac{dy}{dt} \end{matrix}\right|_{t=0}=0 . The position (accurate to two decimal places) of the particle at t = \pi is _______.
\frac{d^{2}y}{dt^{2}}=-\frac{dy}{dt}-\frac{5y}{4}
The initial conditions are y(0) = 1 and \left.\begin{matrix} \frac{dy}{dt} \end{matrix}\right|_{t=0}=0 . The position (accurate to two decimal places) of the particle at t = \pi is _______.
-0.21 | |
-0.5 | |
-0.01 | |
-0.9 |
Question 7 Explanation:
\begin{aligned} \frac{d^{2}}{d t^{2}}+\frac{dy}{dt}+\frac{5Y}{4}&=0\\ y(0)&=1 \\ y^{\prime}(0)&=0 \end{aligned}
By applying Laplace transform,
\begin{aligned} s^{2} Y(s)-s(1)+s Y(s)-1+\frac{5}{4} Y(s)=0 & \\ Y(s)=\frac{s+1}{s^{2}+s+\frac{5}{4}}=\frac{s+1}{\left(s+\frac{1}{2}\right)^{2}+1} \\ =\frac{\left(s+\frac{1}{2}\right)}{\left(s+\frac{1}{2}\right)^{2}+1}+\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^{2}+1} \end{aligned}
By taking inverse Laplace transform.
\begin{aligned} y(t)&=e^{-t / 2}\left[\cos (t)+\frac{1}{2} \sin (t)\right] ; t>0\\ \text{At }t=\pi &\\ y(t=\pi) &=e^{-\pi / 2}[(-1)+(0)] \\ &=-e^{-\pi / 2}=-0.2078 \simeq-0.21 \end{aligned}
By applying Laplace transform,
\begin{aligned} s^{2} Y(s)-s(1)+s Y(s)-1+\frac{5}{4} Y(s)=0 & \\ Y(s)=\frac{s+1}{s^{2}+s+\frac{5}{4}}=\frac{s+1}{\left(s+\frac{1}{2}\right)^{2}+1} \\ =\frac{\left(s+\frac{1}{2}\right)}{\left(s+\frac{1}{2}\right)^{2}+1}+\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^{2}+1} \end{aligned}
By taking inverse Laplace transform.
\begin{aligned} y(t)&=e^{-t / 2}\left[\cos (t)+\frac{1}{2} \sin (t)\right] ; t>0\\ \text{At }t=\pi &\\ y(t=\pi) &=e^{-\pi / 2}[(-1)+(0)] \\ &=-e^{-\pi / 2}=-0.2078 \simeq-0.21 \end{aligned}
Question 8 |
The general solution of the differential equation \frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}-5y=0 in terms of arbitrary constants K_{1} \; and \; K_{2} is
K_{1}e^{(-1+\sqrt{6})x}+K_{2}e^{(-1-\sqrt{6})x} | |
K_{1}e^{(-1+\sqrt{8})x}+K_{2}e^{(-1-\sqrt{8})x} | |
K_{1}e^{(-2+\sqrt{6})x}+K_{2}e^{(-2-\sqrt{6})x} | |
K_{1}e^{(-2+\sqrt{8})x}+K_{2}e^{(-2-\sqrt{8})x} |
Question 8 Explanation:
The general solution of the differential equation
\begin{aligned} \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-5 y&=0 \\ \left(D^{2}+2 D-5\right) y&=0 \end{aligned}
Auxillary equation is
\begin{aligned} m^{2}+2 m-5 &=0 \\ m &=\frac{-2 \pm \sqrt{4+20}}{2}=\frac{-2 \pm \sqrt{24}}{2} \\ &=\frac{-2 \pm 2 \sqrt{6}}{2}=-1 \pm \sqrt{6} \\ \text { Solution is } \quad y &=K_{1} e^{(-1+\sqrt{6}) x}+K_{2} e^{(-1-\sqrt{6}) x} \end{aligned}
\begin{aligned} \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-5 y&=0 \\ \left(D^{2}+2 D-5\right) y&=0 \end{aligned}
Auxillary equation is
\begin{aligned} m^{2}+2 m-5 &=0 \\ m &=\frac{-2 \pm \sqrt{4+20}}{2}=\frac{-2 \pm \sqrt{24}}{2} \\ &=\frac{-2 \pm 2 \sqrt{6}}{2}=-1 \pm \sqrt{6} \\ \text { Solution is } \quad y &=K_{1} e^{(-1+\sqrt{6}) x}+K_{2} e^{(-1-\sqrt{6}) x} \end{aligned}
Question 9 |
Which one of the following is the general solution of the first order differential equation \frac{dy}{dx}=(x+y-1)^{2} , where x, y are real?
y=1+x+tan^{-1}(x+c), where c is a constant | |
y=1+x+tan(x+c), where c is a constant | |
y=1-x+tan^{-1}(x+c), where c is a constant | |
y=1-x+tan(x+c), where c is a constant |
Question 9 Explanation:
\begin{aligned} \frac{d y}{d x}&=(x+y-1)^{2} &\ldots(i)\\ \text{Let, }x+y-1&=t &\ldots(ii)\\ 1+\frac{d y}{d x} &=\frac{d t}{d x} \\ \frac{d y}{d x} &=\frac{d t}{d x}-1 &\ldots(iii) \end{aligned}
Substituting equations (ii) and (iii) in equation (i)
\begin{aligned} \frac{d t}{d x}-1&=t^{2} \\ \frac{d t}{t^{2}+1}&=d x \end{aligned}
Integrating both side
\begin{aligned} \int \frac{1}{t^{2}+1} d t &=\int d x \\ \tan ^{-1} t &=x+c\\ \text{Since, }\quad\quad t&=x+y-1 \\ \therefore \tan ^{-1}(x+y-1)&=x+c \\ x+y-1&=\tan (x+c) \\ y&=1-x+\tan (x+c) \end{aligned}
Substituting equations (ii) and (iii) in equation (i)
\begin{aligned} \frac{d t}{d x}-1&=t^{2} \\ \frac{d t}{t^{2}+1}&=d x \end{aligned}
Integrating both side
\begin{aligned} \int \frac{1}{t^{2}+1} d t &=\int d x \\ \tan ^{-1} t &=x+c\\ \text{Since, }\quad\quad t&=x+y-1 \\ \therefore \tan ^{-1}(x+y-1)&=x+c \\ x+y-1&=\tan (x+c) \\ y&=1-x+\tan (x+c) \end{aligned}
Question 10 |
The particular solution of the initial value problem given below is
\frac{d^{2}y}{dx^{2}}+12\frac{dy}{dx}+36y=0 with y(0)=3 and \frac{dy}{dx}|_{x=0}=-36
\frac{d^{2}y}{dx^{2}}+12\frac{dy}{dx}+36y=0 with y(0)=3 and \frac{dy}{dx}|_{x=0}=-36
(3-18x)e^{-6x} | |
(3+25x)e^{-6x} | |
(3+20x)e^{-6x} | |
(3-12x)e^{-6x} |
Question 10 Explanation:
\begin{aligned} \left(D^{2}+12 D+36\right) &=0 \\ (D+6)^{2} y &=0 \\ D &=-6,-6\\ y &=\left(C_{1}+x C_{2}\right) e^{6 x} \\ y &=C_{1} e^{-6 x}+C_{2} x e^{-6 x} \\ y(0) &=3 \\ &=C_{1}+0 \\ \Rightarrow \quad C_{1} &=3 \\ y^{\prime} &=-6 C_{1} e^{6 x}+C_{2} \sigma^{6 x}-6 C_{2} x \theta^{6} x \\ y(0) &=-36 \\ -36 &=-6 C_{1}+C_{2} \\ -36 &=-18+C_{2} \\ C_{2} &=-18 \\ \therefore \quad \quad y &=3 e^{-6 x}-18 x \mathrm{e}^{-6 x} \\ y &=(3-18 x) e^{-6 x} \end{aligned}
There are 10 questions to complete.