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}
There are 5 questions to complete.