GATE ECE 2020 | Differential Equations Question 6 | Engineering Mathematics

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GATE ECE 2020

MCQ (Single Correct Answer)

-0.33

The general solution of $\frac{d^2 y}{d x^2}-6 \frac{d y}{d x}+9 y=0$ is

$y=C_1 e^{3 x}+C_2 e^{-3 x}$

$y=C_1 e^{3 x}$

$y=\left(C_1+C_2 x\right) e^{3 x}$

$y=\left(C_1+C_2 x\right) e^{-3 x}$

GATE ECE 2017 Set 2

MCQ (Single Correct Answer)

-0.3

The general solution of the differential equation $$\,\,{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} - 5y = 0\,\,\,$$ in terms of arbitrary constants $${K_1}$$ and $${K_2}$$ is

$${K_1}{e^{\left( { - 1 + \sqrt 6 } \right)x}} + {K_2}{e^{\left( { - 1 - \sqrt 6 } \right)x}}$$

$${K_1}{e^{\left( { - 1 + \sqrt 8 } \right)x}} + {K_2}{e^{\left( { - 1 - \sqrt 8 } \right)x}}$$

$${K_1}{e^{\left( { - 2 + \sqrt 6 } \right)x}} + {K_2}{e^{\left( { - 2 - \sqrt 6 } \right)x}}$$

$${K_1}{e^{\left( { - 2 + \sqrt 8 } \right)x}} + {K_2}{e^{\left( { - 2 - \sqrt 8 } \right)x}}$$

GATE ECE 2017 Set 1

MCQ (Single Correct Answer)

-0.3

Consider the following statements about the linear dependence of the real valued functions $${y_1} = 1,\,\,{y_2} = x$$ and $${y_3} = {x^2}$$. Over the field of real numbers.

$${\rm I}.\,\,\,\,\,$$ $${y_1},{y_2}$$ and $${y_3}$$ are linearly independent on $$ - 1 \le x \le 0$$
$${\rm II}.\,\,\,\,\,$$ $${y_1},{y_2}$$ and $${y_3}$$ are linearly dependent on $$0 \le x \le 1$$
$${\rm III}.\,\,\,\,\,$$ $${y_1},{y_2}$$ and $${y_3}$$ are linearly independent on $$0 \le x \le 1$$
$${\rm IV}.\,\,\,\,\,$$ $${y_1},{y_2}$$ and $${y_3}$$ are linearly dependent on $$ - 1 \le x \le 0$$

Which one among the following is correct?

Both $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ are true

Both $${\rm I}$$ and $${\rm I}$$$${\rm I}$$$${\rm I}$$ are true

Both $${\rm I}$$$${\rm I}$$ and $${\rm IV}$$ are true

Both $${\rm I}$$$${\rm I}$$$${\rm I}$$ and $${\rm IV}$$ are true

GATE ECE 2015 Set 2

MCQ (Single Correct Answer)

-0.3

Consider the differential equation $$\,\,{{dx} \over {dt}} = 10 - 0.2\,x$$ with initial condition $$x(0)=1.$$ The response $$x(t)$$ for $$t > 0$$ is

$$2 - {e^{ - 0.2t}}$$

$$2 - {e^{ 0.2t}}$$

$$50 - 49\,{e^{ - 0.2t}}$$