GATE ECE 2020 | Differential Equations Question 7 | Engineering Mathematics

GATE ECE 2020

MCQ (Single Correct Answer)

-0.67

Which one of the following options contains two solutions of the differential equation $\frac{d y}{d x}=(y-1) x$ ?

$\ln |y-1|=0.5 x^2+c$ and $y=-1$

$\ln |y-1|=2 x^2+c$ and $y=-1$

$\ln |y-1|=0.5 x^2+c$ and $y=1$

$\ln |y-1|=2 x^2+c$ and $y=1$

GATE ECE 2018

Numerical

-0

The position of a particle y(t) is described by the differential equation :

$${{{d^2}y} \over {d{t^2}}} = - {{dy} \over {dt}} - {{5y} \over 4}$$.

The initial conditions are y(0) = 1 and $${\left. {{{dy} \over {dt}}} \right|_{t = 0}}$$ = 0.

The position (accurate to two decimal places) of the particle at t = $$\pi $$ is _______.

Your input ____

GATE ECE 2018

MCQ (Single Correct Answer)

-0.67

A curve passes through the point ($$x$$ = 1, $$y$$ = 0) and satisfies the differential equation

$${{dy} \over {dx}} = {{{x^2} + {y^2}} \over {2y}} + {y \over x}$$. The equation that describes the curve is

$$\ln \left( {1 + {{{y^2}} \over {{x^2}}}} \right) = x - 1$$

$${1 \over 2}\ln \left( {1 + {{{y^2}} \over {{x^2}}}} \right) = x - 1$$

$${1 \over 2}\ln \left( {1 + {y \over x}} \right) = x - 1$$

$$\ln \left( {1 + {y \over x}} \right) = x - 1$$

GATE ECE 2017 Set 1

MCQ (Single Correct Answer)

-0.6

Which one of the following is the general solution of the first order differential equation $${{dy} \over {dx}} = {\left( {x + y - 1} \right)^2}$$ , where $$x,$$ $$y$$ are real ?

$$y = 1 + x + {\tan ^{ - 1}}\left( {x + c} \right),$$ where $$c$$ is a constant

$$y = 1 + x + \tan \left( {x + c} \right),$$ where $$c$$ is a constant

$$y = 1 - x + {\tan ^{ - 1}}\left( {x + c} \right),$$ where $$c$$ is a constant

$$y = 1 - x + \tan \left( {x + c} \right),$$ where $$c$$ is a constant