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 ?
\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 ?
1 and 2 | |
-1 and -2 | |
2 and 3 | |
-2 and -3 |
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.
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).
\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).
-0.125 | |
0.125 | |
-0.182 | |
0.182 |
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}
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).
0.446 | |
0.124 | |
0.745 | |
0.852 |
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}
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)
\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)
5.4 | |
6.4 | |
2.8 | |
4.2 |
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}
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
\frac{d^3y}{dx^3}+x\left ( \frac{dy}{dx} \right )^{\frac{3}{2}}+x^2y=0
the correct description is
an ordinary differential equation of order 3 and degree 2. | |
an ordinary differential equation of order 3 and degree 3. | |
an ordinary differential equation of order 2 and degree 3. | |
an ordinary differential equation of order 3 and degree 3/2. |
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
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.